UBC Wiki - User contributions [en]

Thread:User talk:NabilFadai/Started working!/reply (3)

2012-10-02T04:21:53Z

NabilFadai: Reply to Started working!

Hi David,

Thanks for the writing style guide! I'll make sure to sure &fnof; and "x" from now on :)

I just checked the Math 100 page again, and there doesn't seem to be any exams under that page. Regrettably, I am only on campus this semester on Tuesdays and Thursdays, so a Friday tutorial won't work for me. However, since there is plenty of other calculus questions to be done in the meantime, I am perfectly fine with just working on those for now :)

Cheers,
Nabil

Thread:User talk:NabilFadai/Started working!/reply

2012-10-01T23:48:01Z

NabilFadai: Reply to Started working!

Hi David!

The Wiki is great. Currently I'm only working on exams that have been partially started as I don't know how to create a new exam page (for example, the Dec 11 exam for Math 100). Would you mind explaining to me how to do that?

Also, the 'f's in the solutions tend to be a little inconsistent in my solutions. The reason for this is that when you write f' in LaTeX, it sometimes goes into its nice large format f'(x) and sometimes stays tiny, depending on what other symbols are in that part of the math script. If you're confused by what I'm talking about, check out my solution to Math 104 2c) and it will be super obvious. Any suggestions on how to fix this?

Cheers,
Nabil

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-10-01T23:42:39Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that f'(-3)=f'(3)>0. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields f'(-2)=f'(2)=-2<0. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math> terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is increasing on the intervals <math>(- \infty,-\sqrt 6 ) \cup (-1,1) \cup (\sqrt6,\infty)</math> and decreasing on the intervals <math>(- \sqrt 6,-\sqrt 3 ) \cup (-\sqrt 3,-1) \cup (1,\sqrt 3) \cup (\sqrt3,\sqrt 6)</math>. Note that the critical points are not included in the intervals of increase and decrease.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-10-01T23:39:46Z

NabilFadai:

To determine where f"(x)=0, we look at the numerator of f"(x) and note that f"(x)=0 when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where f"(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f"(x) does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f"(x)</math> at that point to determine if f"(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-10-01T23:38:23Z

NabilFadai:

To determine where f" f<math>''(x)=0</math>, we look at the numerator of f<math>''(x)</math> and note that f<math>''(x)=0</math> when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where f<math>''(x)</math> does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f<math>''(x)</math> does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f<math>''(x)</math> at that point to determine if f<math>''(x)</math> did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-10-01T23:37:34Z

NabilFadai:

To determine where f<math>''(x)=0</math>, we look at the numerator of f<math>''(x)</math> and note that f<math>''(x)=0</math> when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where f<math>''(x)</math> does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f<math>''(x)</math> does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f<math>''(x)</math> at that point to determine if f<math>''(x)</math> did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-10-01T23:37:06Z

NabilFadai:

To determine where f<math>''(x)=0</math>, we look at the numerator of f<math>''(x)</math> and note that f<math>''(x)=0</math> when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where f<math>''(x)</math> does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f<math>''(x)</math> does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f<math>''(x)</math> at that point to determine if <math>f''(x)</math> did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)

2012-09-29T19:16:17Z

NabilFadai:

[[Category:MER RQ flag]][[Category:MER RH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T19:14:57Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math> terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is increasing on the intervals <math>(- \infty,-\sqrt 6 ) \cup (-1,1) \cup (\sqrt6,\infty)</math> and decreasing on the intervals <math>(- \sqrt 6,-\sqrt 3 ) \cup (-\sqrt 3,-1) \cup (1,\sqrt 3) \cup (\sqrt3,\sqrt 6)</math>. Note that the critical points are not included in the intervals of increase and decrease.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T19:13:53Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is increasing on the intervals <math>(- \infty,-\sqrt 6 ) \cup (-1,1) \cup (\sqrt6,\infty)</math> and decreasing on the intervals <math>(- \sqrt 6,-\sqrt 3 ) \cup (-\sqrt 3,-1) \cup (1,\sqrt 3) \cup (\sqrt3,\sqrt 6)</math>. Note that the critical points are not included in the intervals of increase and decrease.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T19:07:22Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is increasing on the intervals <math>(- \infty,-\sqrt 6 ) \cup</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T19:05:13Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is increasing on the intervals <math>(- 5,-\sqrt 6 )<\math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T19:02:10Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is increasing on the intervals <math>(- 5,-\sqrt 6 ) \cup<\math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T18:59:25Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is decreasing on the intevals <math>(-\infty,-\sqrt 6 )<\math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T18:58:55Z

NabilFadai:

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}>0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is decreasing on the intevals <math>(-\inf,-\sqrt 6 )<\math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Solution 1

2012-09-29T18:53:31Z

NabilFadai: Created page with "From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>. Testing a value less than ..."

From Question 2 (a), we know that the critical points of f(x), when f'(x)=0 or does not exist, are at <math>x=\pm1,\pm\sqrt 6, \pm \sqrt 3</math>.

Testing a value less than <math>-\sqrt 6</math>, such as -3, we get <math>f'(-3)=\frac{2}{3}>0 </math>. Note also that <math>f'(-3)=f'(3)>0 </math>. Testing a point between <math>-\sqrt 6</math> and <math>-\sqrt 3</math>, such as -2, yields <math>f'(-2)=f'(2)=-2<0</math>. Testing a point between <math>-\sqrt 3</math> and -1, such as -1.5, we get <math>f'(-1.5)=f'(1.5)=-\frac{25}{3}<0</math>. Finally, we test a point between -1 and 1, such as 0, and get <math>f'(0)=\frac{2}{3}<0</math>.

Normally, we would have to test points between the positive critical points as well, but since our function only has <math>x^2</math>terms in it, we know that our negative test points will have identical values to their corresponding positive test points. For example, testing -2 and 2 produce identical values of -2, so f'(x)<0 between <math>-\sqrt 6</math> and <math>-\sqrt 3</math> as well as between <math>\sqrt 3</math> and <math>\sqrt 6</math>.

f(x) is defined to be increasing when f'(x)>0 and decreasing when f'(x)<0. So, f(x) is decreasing on the intevals <math>(-\infty,-\sqrt 6 )<\math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Hint 1

2012-09-29T18:36:22Z

NabilFadai: Created page with "In question 2 (a), we determined where f'(x)=0 or does not exist, i.e. the critical points of f(x). Try testing values between two adjacent critical points."

In question 2 (a), we determined where f'(x)=0 or does not exist, i.e. the critical points of f(x). Try testing values between two adjacent critical points.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Statement

2012-09-29T18:32:50Z

NabilFadai:

Consider the function

:<math>f(x) = \frac{x^3+x^2-2x-3}{x^2-3}.</math>

Its first and second derivatives are given by

:<math>f'(x) = \frac{(x^2-1)(x^2-6)}{(x^2-3)^2}, \qquad f''(x) = \frac{2x(x^2+9)}{(x^2-3)^3}.</math>

On what intervals is f(x) increasing? On what intervals is f(x) decreasing?

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (c)/Statement

2012-09-29T18:32:01Z

NabilFadai: Created page with "Consider the function :<math>f(x) = \frac{x^3+x^2-2x-3}{x^2-3}.</math> Its first and second derivatives are given by :<math>f'(x) = \frac{(x^2-1)(x^2-6)}{(x^2-3)^2}, \qquad..."

Consider the function

:<math>f(x) = \frac{x^3+x^2-2x-3}{x^2-3}.</math>

Its first and second derivatives are given by

:<math>f'(x) = \frac{(x^2-1)(x^2-6)}{(x^2-3)^2}, \qquad f''(x) = \frac{2x(x^2+9)}{(x^2-3)^3}.</math>

Find all ''x'' such that f'(x) = 0 or f'(x) does not exist.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)

2012-09-29T18:28:27Z

NabilFadai:

[[Category:MER QGQ flag]][[Category:MER RH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-09-29T18:25:29Z

NabilFadai:

To determine where <math>f''(x)=0</math>, we look at the numerator of <math>f''(x)</math> and note that <math>f''(x)=0</math> when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where <math>f''(x)</math> does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, <math>f''(x)</math> does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of <math>f''(x)</math> at that point to determine if <math>f''(x)</math> did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-09-29T18:24:34Z

NabilFadai:

To determine where <math>f''(x)=0</math>, we look at the numerator of <math>f''(x)</math> and note that <math>f''(x)=0</math> when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where <math>f''(x)</math> does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, <math>f''(x)</math> does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f''(x) at that point to determine if f'(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-09-29T18:24:03Z

NabilFadai:

To determine where <math>f''(x)=0</math>, we look at the numerator of f''(x) and note that f''(x)=0 when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where <math>f''(x)</math> does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, <math>f''(x)</math> does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f''(x) at that point to determine if f'(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-09-29T18:22:48Z

NabilFadai:

To determine where f''(x)=0, we look at the numerator of f''(x) and note that f''(x)=0 when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9=0</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where f''(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f''(x) does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f''(x) at that point to determine if f'(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-09-29T18:22:23Z

NabilFadai:

To determine where f''(x)=0, we look at the numerator of f''(x) and note that f''(x)=0 when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since <math>x^2+9</math> has no real solutions, the only value of x that satisfies the above equation is when <math> x=0</math>.

To determine where f''(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f''(x) does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f''(x) at that point to determine if f'(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Solution 1

2012-09-29T18:21:41Z

NabilFadai: Created page with "To determine where f''(x)=0, we look at the numerator of f''(x) and note that f''(x)=0 when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>. Since x^2+9 has no re..."

To determine where f''(x)=0, we look at the numerator of f''(x) and note that f''(x)=0 when the numerator is zero, i.e. <math>\qquad 2x(x^2+9)=0</math>.
Since x^2+9 has no real solutions, the only value of x that satisfies the above equation is when x=0.

To determine where f''(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f''(x) does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f''(x) at that point to determine if f'(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Hint 1

2012-09-29T18:18:29Z

NabilFadai: Created page with "Try setting the numerator and the denominator to 0. What does a value of 0 in the numerator correspond to? How about when the denominator is 0?"

Try setting the numerator and the denominator to 0. What does a value of 0 in the numerator correspond to? How about when the denominator is 0?

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:17:09Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f'(x) and note that f'(x)=0 when the numerator is zero, i.e. <math>\qquad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f'(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>.

Note that the denominator and numerator are never simultaneously 0. If they were, we would have to take the limit of f'(x) at that point to determine if f'(x) did not exist or was zero (or another finite number).

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:15:03Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f'(x) and note that <math>f'(x)=0 </math> when the numerator is zero, i.e. <math>\qquad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f'(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)

2012-09-29T18:13:39Z

NabilFadai:

[[Category:MER QGQ flag]][[Category:MER RH flag]][[Category:MER RS flag]][[Category:MER CT flag]]







{{MER Question page}}

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Hint 1

2012-09-29T18:12:34Z

NabilFadai: Created page with "Try setting the numerator and the denominator to 0. What does a value of 0 in the numerator correspond to? How about when the denominator is 0?"

Try setting the numerator and the denominator to 0. What does a value of 0 in the numerator correspond to? How about when the denominator is 0?

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:08:37Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f'(x) and note that <math>f'(x)=0 </math> when the numerator is zero, i.e.
<math>\qquad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f'(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:08:08Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when the numerator is zero, i.e.
<math>\qquad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:07:04Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when
<math>\qquad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:06:44Z

NabilFadai:

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:06:25Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math>\qquad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:06:07Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math>\quad (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:05:46Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math> (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\quad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

To determine where f'(x) does not exist, we set the denominator to 0.
:<math>\implies x^2-3=0</math>
So, f'(x) does not exist when <math>x=\pm\sqrt 3</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:03:53Z

NabilFadai:

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:03:33Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math> (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\qquad x^2-6=0 </math>
Solving each equation and noting that both positive and negative solutions are valid, we obtain
:<math> f(x)=0 \implies x=\pm1,\pm\sqrt 6</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:03:01Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math> (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0,\qquad x^2-6=0 </math>
Solving each equation and noting that positive and negative solutions are valid, we obtain
:<math> x=\pm1,\pm\sqrt 6</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:02:43Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math> (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0 x^2-6=0 </math>
Solving each equation and noting that positive and negative solutions are valid, we obtain
:<math> x=\pm1,\pm\sqrt 6</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:02:34Z

NabilFadai:

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T18:00:44Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0 </math> when<math> (x^2-1)(x^2-6)=0</math>.
:<math> \implies x^2-1=0 OR x^2-6=0 </math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T17:58:37Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that
:<math>f(x)=0 \implies (x^2-1)(x^2-6)=0</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T17:57:58Z

NabilFadai:

To determine where f'(x)=0, we look at the numerator of f(x) and note that
:<math>f(x)=0</math> when <math>(x^2-1)(x^2-6)=0</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (a)/Solution 1

2012-09-29T17:57:25Z

NabilFadai: Created page with "To determine where f'(x)=0, we look at the numerator of f(x) and note that <math>f(x)=0</math> when <math>(x^2-1)(x^2-6)=0</math>"

To determine where f'(x)=0, we look at the numerator of f(x) and note that
<math>f(x)=0</math> when <math>(x^2-1)(x^2-6)=0</math>

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Statement

2012-09-27T23:43:37Z

NabilFadai:

Consider the function
:<math>
f(x) = \frac{x^3 +x^2-2x-3}{x^2-3}
</math>

Its first and second derivatives are given by
:<math>
f′(x) = \frac{(x^2 − 1)(x^2 − 6)}{[x^2-3)^2}
</math>

Find all x such that f′′(x) = 0 or f′′(x) does not exist.

Science:Math Exam Resources/Courses/MATH104/December 2011/Question 02 (b)/Statement

2012-09-27T23:43:19Z

NabilFadai:

Consider the function
:<math>
f(x) = \frac{x^3 +x^2−2x−3}{x^2−3}
</math>

Its first and second derivatives are given by
:<math>
f′(x) = \frac{(x^2 − 1)(x^2 − 6)}{[x^2-3)^2}
</math>

Find all x such that f′′(x) = 0 or f′′(x) does not exist.

User:NabilFadai

2012-09-27T23:41:20Z

NabilFadai: Created page with "Category: MER Participant"

[[Category: MER Participant]]