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Science:Math Exam Resources/Courses/MATH100/December 2018/Question 06/Hint 1

2020-05-01T10:51:00Z

DANIELDIBENEDETTO:

Use Pythagoras' theorem to relate the positions of the two particles.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 06/Solution 1

2020-05-01T10:49:54Z

DANIELDIBENEDETTO:

Let <math> x </math> be the distance from the origin to the particle <math> A </math> and let <math> y </math> be the distance from the origin to the particle <math>B </math>. Also Let <math> a </math> be the distance between <math> A </math> and <math> B </math>. Notice that <math> x,y </math> and <math>a </math> are functions of time and we shall thus write them as <math> x(t),y(t),a(t) </math> respectively.

By the given information, we have <math> \frac{dx}{dt}=2 </math> units/min and <math> \frac{dy}{dt}=-1</math> units/min. Using the initial positions, then we see that <math>x(t)=4+2t</math> and <math>y(t)=8-t</math>. We can relate these two functions by Pythagoras' theorem: we have <math> x(t)^2+y(t)^2=a(t)^2 </math>, and therefore

<math>a(t)^2=x(t)^2+y(t)^2=(4+2t)^2+(8-t)^2=16+16t+4t^2+64-16t+t^2=80+5t^2.</math>

Now, we will first determine the time <math>t</math> at which the distance between the particles is <math>10</math> units. We do this by setting <math>a(t)=10</math> and solving for <math>t</math>. This gives

<math> 100=80+5t^2</math>

and therefore <math>t=2</math> (we can ignore the negative root since time cannot be negative).

Next, we want to find <math>a'(t) </math> when <math>a(t)=10</math>. By differentiating both sides of the equation <math>a(t)^2=80+5t^2</math>, we get <math>2a(t)a'(t)=10t</math>. Now we can plug in <math>a(t)=10</math> and <math>t=2</math>, which gives <math>\color{blue}a'(2)=1 \text{ unit per min}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 07/Solution 1

2020-04-30T16:43:30Z

DANIELDIBENEDETTO:

We are being asked to show that the global maximum of <math>f(x)=\sin(x)+\sqrt{3}\cos(x)</math> is at most equal to <math>2</math> and its global minimum is at least equal to <math>-2</math>. For this, note that the function is periodic with period <math>2\pi</math> and therefore suffices to find the global maximum and the global minimum on the closed interval <math>[0,2\pi]</math>

First determine the critical points in the interval by computing the derivative and setting it equal to 0:

<math>f'(x)=\cos(x)-\sqrt{3}\cdot \sin(x),</math>

so <math> f'(x)=0</math> implies that <math>\tan(x)=\frac{1}{\sqrt{3}}</math>. The only solutions of thid equation in the interval <math>(0,2\pi)</math> are <math>x=\frac{\pi}{6}</math> and <math>x=\pi+\frac{\pi}{6}</math>. It remains to check the value of <math>f(x)</math> at the endpoints of the interval and at the two critical points:

<math>f(0)=\sqrt{3}\text{, }f\left(\frac{\pi}{6}\right)=\frac{1}{2}+\sqrt{3}\cdot \frac{\sqrt{3}}{2}=2.</math>

<math>f\left(\pi+\frac{\pi}{6}\right)=-\frac{1}{2}+\sqrt{3}\cdot \frac{-\sqrt{3}}{2}=-2\text{ and }f(2\pi)=\sqrt{3}.</math>

Hence, <math>\color{blue}\text{the global maximum is } 2 \text{ and the global minimum is } -2</math>, as desired.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 07/Hint 1

2020-04-30T16:38:15Z

DANIELDIBENEDETTO:

Find the absolute maximum and minimum of the function.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Solution 1

2020-04-30T14:59:08Z

DANIELDIBENEDETTO:

We know that <math>f</math> is at least three times differentiable, so we will use the second degree Maclaurin polynomial, which is

<math>A=T_2(x)=f(0)+f'(0)x+\frac{f''(0)}{2}\cdot x^2=1-2x+\frac{3x^2}{2}.</math>

That is, <math>f(2)</math> is approximated by <math>\color{blue}T(2)=1-2\cdot 2+\frac{3\cdot 2^2}{2}=3</math>.

To estimate the error of our approximation, by the Lagrange Remainder Theorem we have

<math>f(2)-T_2(2)=\frac{f'''(c)}{3!}\cdot (2-0)^3=\frac{\frac{6-c^2}{9-\cos(c)}}{6}\cdot 8=\frac{4(6-c^2)}{3(9-\cos(c))},</math>

where <math>c</math> is in the interval <math>[0,2]</math>. Note that this expression is always positive for <math>c</math> in <math>[0,2]</math>. Furthermore, the numerator is decreasing on this interval and the denominator attains its minimum at <math>c=0</math>, which implies that this error is maximised at <math>c=0</math>. We conclude that the error <math>\color{blue}|f(2)-T_2(2)|\le \frac{4\cdot 6}{3\cdot 8} = 1</math>, as required.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Hint 2

2020-04-30T14:50:04Z

DANIELDIBENEDETTO:

Use the Lagrange Remainder Theorem to estimate the error of your approximation.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Hint 1

2020-04-30T14:49:20Z

DANIELDIBENEDETTO:

Use the Maclaurin polynomial.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Hint 1

2020-04-30T14:48:24Z

DANIELDIBENEDETTO:

Use Taylor's Formula centered at a nearby point.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Statement

2020-04-30T14:47:24Z

DANIELDIBENEDETTO:

Let <math>f(x)</math> be a differentiable function satisfying the following properties:
<math> f(0) = 1, f'(0) = -2, f''(0) = 3 ;</math> and
<math> f'''(x) = \frac{6 - x^2}{9 - \cos(x)} </math>
Determine with proof an approximation <math>A</math> to <math>f(2)</math> with the property that the error in the approximation is at most <math>1</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 08/Statement

2020-04-30T14:47:13Z

DANIELDIBENEDETTO:

Let <math>f(x)</math> be a differentiable function satisfying the following properties:
<math> f(0) = 1, f'(0) = -2, f''(0) = 3 ;</math> and
<math> f'''(x) = \frac{6 - x^2}{9 - \cos(x)} </math>
Determine with proof and approximation <math>A</math> to <math>f(2)</math> with the property that the error in the approximation is at most <math>1</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Hint 2

2020-04-30T14:23:40Z

DANIELDIBENEDETTO: Created page with "Remember to check the necessary conditions for the theorem you want to apply."

Remember to check the necessary conditions for the theorem you want to apply.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Hint 1

2020-04-30T14:22:58Z

DANIELDIBENEDETTO:

Start by letting <math> h(x)=f(x)-\frac{x+1}{2}=g(x)+\sin(x)-\frac{x+1}{2} </math>. Find two numbers <math> c_1 </math> and <math> c_2 </math> such that <math> h(c_1)>0 </math> and <math> h(c_2)<0 </math>. Which theorem can we use now?

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Solution 1

2020-04-30T14:20:12Z

DANIELDIBENEDETTO:

Let <math> h(x)=f(x)-\frac{x+1}{2}=g(x)+\sin(x)-\frac{x+1}{2} </math>. Note that since <math> g</math> is differentiable (and since <math> \sin(x)</math> is differentiable), then <math> h</math> is also differentiable (and therefore continuous).

Let <math> c_1=\frac{\pi}{2} </math>, then <math> h(c_1)=g(c_1)+\sin(c_1)-\frac{c_1+1}{2}\geq \frac{c_1}{2}+\sin(c_1)-\frac{c_1+1}{2}=\sin(c_1)-\frac{1}{2}=\frac{1}{2}>0 </math>.

Let <math> c_2=-\frac{\pi}{2} </math>, then <math> h(c_2)=g(c_2)+\sin(c_2)-\frac{c_2+1}{2}\leq \frac{c_2}{2}+1+\sin(c_2)-\frac{c_2+1}{2}=\sin(c_2)+\frac{1}{2}=-\frac{1}{2}<0 </math>.

Thus, by the intermediate value theorem (which applies because, as we saw, <math> h</math> is continuous), there exists <math>c\in(-\frac{\pi}{2},\frac{\pi}{2}) </math> such that <math> h(c)=0 </math>.

Using the same method, by the periodicity of <math> \sin(x)</math> we see that for any integer <math> n </math>, there exists <math> c\in (-\frac{\pi}{2}+2n\pi,\frac{\pi}{2}+2n\pi) </math> such that <math> h(c)=0 </math>. <math>\color{blue} \text{So there are infinitely many real numbers } c \text{ such that }f(c)=\frac{c+1}{2} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Solution 1

2020-04-30T14:19:58Z

DANIELDIBENEDETTO:

Let <math> h(x)=f(x)-\frac{x+1}{2}=g(x)+\sin(x)-\frac{x+1}{2} </math>. Note that since <math> g</math> is differentiable (and since <math> \sin(x)</math> is differentiable), then <math> h</math> is also differentiable (and therefore continuous).

Let <math> c_1=\frac{\pi}{2} </math>, then <math> h(c_1)=g(c_1)+\sin(c_1)-\frac{c_1+1}{2}\geq \frac{c_1}{2}+\sin(c_1)-\frac{c_1+1}{2}=\sin(c_1)-\frac{1}{2}=\frac{1}{2}>0 </math>.

Let <math> c_2=-\frac{\pi}{2} </math>, then <math> h(c_2)=g(c_2)+\sin(c_2)-\frac{c_2+1}{2}\leq \frac{c_2}{2}+1+\sin(c_2)-\frac{c_2+1}{2}=\sin(c_2)+\frac{1}{2}=-\frac{1}{2}<0 </math>.

Thus, by the intermediate value theorem (which applies because, as we saw, <math> h</math> is continuous), there exists <math>c\in(-\frac{\pi}{2},\frac{\pi}{2}) </math> such that <math> h(c)=0 </math>.

Using the same method, by the periodicity of <math> \sin(x)</math> we see that for any integer <math> n </math>, there exists <math> c\in (-\frac{\pi}{2}+2n\pi,\frac{\pi}{2}+2n\pi) </math> such that <math> h(c)=0 </math>. <math>\color{blue} \text{So there are infinite real numbers } c \text{ such that }f(c)=\frac{c+1}{2} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Solution 1

2020-04-30T14:19:47Z

DANIELDIBENEDETTO:

Let <math> h(x)=f(x)-\frac{x+1}{2}=g(x)+\sin(x)-\frac{x+1}{2} </math>. Note that since <math> g</math> is differentiable (and since <math> \sin(x)</math> is differentiable), then <math> h</math> is also differentiable (and therefore continuous).

Let <math> c_1=\frac{\pi}{2} </math>, then <math> h(c_1)=g(c_1)+\sin(c_1)-\frac{c_1+1}{2}\geq \frac{c_1}{2}+\sin(c_1)-\frac{c_1+1}{2}=\sin(c_1)-\frac{1}{2}=\frac{1}{2}>0 </math>.

Let <math> c_2=-\frac{\pi}{2} </math>, then <math> h(c_2)=g(c_2)+\sin(c_2)-\frac{c_2+1}{2}\leq \frac{c_2}{2}+1+\sin(c_2)-\frac{c_2+1}{2}=\sin(c_2)+\frac{1}{2}=-\frac{1}{2}<0 </math>.

Thus, by the intermediate value theorem (which applies because, as we saw, <math> h</math> is continuous), there exists <math>c\in(-\frac{\pi}{2},\frac{\pi}{2}) </math> such that <math> h(c)=0 </math>.

Using the same method, by the periodicity of <math> \sin(x)</math> we see that for any integer <math> n </math>, there exists <math> c\in (-\frac{\pi}{2}+2n\pi,\frac{\pi}{2}+2n\pi) </math> such that <math> h(c)=0 </math>. <math>\color{blue} \text{So there are infinite real numbers } c { such that }f(c)=\frac{c+1}{2} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 11 (a)/Solution 1

2020-04-30T14:19:09Z

DANIELDIBENEDETTO:

Let <math> h(x)=f(x)-\frac{x+1}{2}=g(x)+\sin(x)-\frac{x+1}{2} </math>. Note that since <math> g</math> is differentiable (and since <math> \sin(x)</math> is differentiable), then <math> h</math> is also differentiable (and therefore continuous).

Let <math> c_1=\frac{\pi}{2} </math>, then <math> h(c_1)=g(c_1)+\sin(c_1)-\frac{c_1+1}{2}\geq \frac{c_1}{2}+\sin(c_1)-\frac{c_1+1}{2}=\sin(c_1)-\frac{1}{2}=\frac{1}{2}>0 </math>.

Let <math> c_2=-\frac{\pi}{2} </math>, then <math> h(c_2)=g(c_2)+\sin(c_2)-\frac{c_2+1}{2}\leq \frac{c_2}{2}+1+\sin(c_2)-\frac{c_2+1}{2}=\sin(c_2)+\frac{1}{2}=-\frac{1}{2}<0 </math>.

Thus, by the intermediate value theorem (which applies because, as we saw, <math> h</math> is continuous), there exists <math>c\in(-\frac{\pi}{2},\frac{\pi}{2}) </math> such that <math> h(c)=0 </math>.

Using the same method, by the periodicity of <math> \sin(x)</math> we see that for any integer <math> n </math>, there exists <math> c\in (-\frac{\pi}{2}+2n\pi,\frac{\pi}{2}+2n\pi) </math> such that <math> h(c)=0 </math>. So there are infinite real numbers <math> c </math> such that <math> f(c)=\frac{c+1}{2} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (d)/Solution 1

2020-04-30T11:29:30Z

DANIELDIBENEDETTO:

We know from part (c) that <math> f''(x) </math> changes sign at <math> x= 5/2 </math>, and we have <math>f(5/2) = (5/2)^{9/7} + (9/2)(5/2)^{2/7}=7\cdot (5/2)^{2/7}</math>. Therefore, the only inflection point is at <math>\color{blue} (x,y)=\left(5/2, 7\cdot (5/2)^{2/7}\right)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (d)/Solution 1

2020-04-30T11:29:22Z

DANIELDIBENEDETTO:

We know from part (c) that <math> f''(x) </math> changes sign at <math> x= 5/2 </math>, and we have <math>f(5/2) = (5/2)^{9/7} + (9/2)(5/2)^{2/7}=7\cdot (5/2)^{2/7}</math>. Therefore, the inflection point is at <math>\color{blue} (x,y)=\left(5/2, 7\cdot (5/2)^{2/7}\right)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (d)/Solution 1

2020-04-30T11:29:02Z

DANIELDIBENEDETTO:

We know from part (c) that <math> f''(x) </math> changes sign at <math> x= 5/2 </math>, and we have <math>f(5/2) = (5/2)^{9/7} + (9/2)(5/2)^{2/7}=7\cdot (5/2)^{2/7}</math>. Therefore, the inflection point is <math>\color{blue} (x,y)=\left(7\cdot (5/2)^{2/7}\right)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (d)/Solution 1

2020-04-30T11:28:33Z

DANIELDIBENEDETTO:

We know from part (c) that <math> f''(x) </math> changes sign at <math> x= 5/2 </math>, and we have <math>f(5/2) = (5/2)^{9/7} + (9/2)(5/2)^{2/7}=7\cdot (5/2)^{2/7}</math>. Therefore, the inflection point is <math>\color{blue} x=5/2, y=7\cdot (5/2)^{2/7}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (d)/Solution 1

2020-04-30T11:25:16Z

DANIELDIBENEDETTO:

We know from part (c) that <math> f''(x) </math> changes sign at <math> x= 5/2 </math>. Therefore, the inflection point is <math>\color{blue} x=5/2, y=(5/2)^{9/7} + (9/2)(5/2)^{2/7}=7\cdot (5/2)^{2/7}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (c)/Solution 1

2020-04-30T11:23:53Z

DANIELDIBENEDETTO:

We have <math> f''(x)=\frac{9}{7}\cdot \frac{1\cdot x^{\frac{5}{7}}-(x+1)\cdot \frac{5}{7x^{\frac{2}{7}}}}{x^{\frac{10}{7}}}= \frac{9}{7}\cdot \frac{7x-5(x+1)}{7x^{\frac{12}{7}}}=\frac{9(2x-5)}{49x^{\frac{12}{7}}}. </math>
This denominator will always be nonnegative (because of the even number <math>12</math> in the exponent), so it suffices to understand when the numerator is positive/negative. Note also that <math> f''(x)</math> is undefined at <math> x=0</math>. The numerator is zero when <math>x=5/2</math>, positive when <math>x>5/2</math>, and negative when <math>x<5/2</math>. We therefore conclude that <math>\color{blue} f(x) \text{ is concave up on } \left(\frac{5}{2}, +\infty\right) \text{ and concave down on }\left(-\infty,0\right)\cup\left(0, \frac{5}{2}\right) </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (a)/Solution 1

2020-04-29T13:52:12Z

DANIELDIBENEDETTO:

We take the derivative <math> f'(x) = \frac{9}{7} (x^{\frac{2}{7}} + x^{-\frac{5}{7}}) = \frac{9}{7}\cdot \frac{x+1}{x^{\frac{5}{7}}}</math>.
The function therefore has two critical points: at <math>x=-1,0</math>. Now using a chart of signs shows that <math>f'(x)</math> exactly when <math>x\in(-1,0)</math>. We conclude that <math>\color{blue}f(x)\text{ is decreasing on }(-1,0) \text{ and increasing on } (-\infty, -1)\cup (0,+\infty)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (b)/Solution 1

2020-04-29T13:50:33Z

DANIELDIBENEDETTO:

We can see from <math> f'(x) = \frac{9}{7}\cdot \frac{x+1}{x^{\frac{5}{7}}} </math> that <math> f'(x) =0 </math> if <math> x = -1 </math> and undefined if <math> x = 0 </math>. We also know <math> f(x) </math> is increasing for <math> x \in (-\infty,-1) </math> and decreasing for <math> x \in (-1,0) </math>. Therefore, a local maximum occurs at <math> x = -1 </math>. Moreover, a local minimum occurs at <math> x = 0 </math> since <math> f(x) </math> is decreasing for <math> x \in (-1,0) </math> and increasing for for <math> x \in (0,\infty) </math>. Hence, <math>\color{blue} f(x) \text{ has a local minimum at } (0,0) \text{ and a local maximum at } \left(-1,\frac{7}{2}\right) </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (b)/Hint 1

2020-04-29T13:47:35Z

DANIELDIBENEDETTO:

What are the critical points? Remember to consider value(s) <math> c </math> where <math> f'(c) </math> is undefined as well.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (a)/Solution 1

2020-04-29T13:35:02Z

DANIELDIBENEDETTO:

We take the derivative <math> f'(x) = \frac{9}{7} (x^{\frac{2}{7}} + x^{-\frac{5}{7}}) = \frac{9}{7}\cdot \frac{x+1}{x^{\frac{5}{7}}}</math>.
This function has two critical points: at <math>x=-1,0</math>. Now using a chart of signs shows that <math>f'(x)</math> exactly when <math>x\in(-1,0)</math>. We conclude that <math>\color{blue}f(x)\text{ is decreasing on }(-1,0) \text{ and increasing on } (-\infty, -1)\cup (0,+\infty)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 05 (a)/Solution 1

2020-04-29T13:34:53Z

DANIELDIBENEDETTO:

We take the derivative <math> f'(x) = \frac{9}{7} (x^{\frac{2}{7}} + x^{-\frac{5}{7}}) = \frac{9}{7}\cdot \frac{x+1}{x^{\frac{5}{7}}}</math>.
This function has two critical points: at <math>x=-1,0</math>. Now using a chart of signs shows that <math>f'(x)</math> exactly when <math>x\in(-1,0)</math>. We conclude that <\math>\color{blue}f(x)\text{ is decreasing on }(-1,0) \text{ and increasing on } (-\infty, -1)\cup (0,+\infty)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 04/Solution 1

2020-04-29T13:28:28Z

DANIELDIBENEDETTO:

Let <math> (x_0, y_0) </math> be a point on the curve. Then this point satisfies

<math> x_0^2 + 2y_0^2 = 8.~~~~(1) </math>

To find the slope of the tangent line passing through <math> (x_0, y_0) </math>, use implicit differentiation. Differentiate both sides of the equation of the curve with respect to <math>x</math>. By the chain rule,

<math> 2x + 4yy' = 0. </math>

Let <math>y_0'</math> be the slope of the tangent line at point <math>(x_0, y_0)</math>. Assume <math>y_0 \neq 0</math>. Then

<math>y'_0 = -\frac{x_0}{2y_0}.</math>

Using the point-slope formula, the tangent line equation to the curve at the point <math>(x_0, y_0)</math> can be written as

<math> y - y_0 = -\frac{x_0}{2y_0}(x - x_0). </math>

Let this tangent line pass through <math>(0, -6)</math>. Putting <math>x = 0</math> and <math>y = -6</math>,

<math> -6 - y_0 = \frac{x_0^2}{2y_0}. </math>

Together with Equation (1) we can solve for the point. <math>(x_0, y_0) = \left(\frac{8}{3}, -\frac{2}{3}\right)</math> or <math>\left(-\frac{8}{3}, -\frac{2}{3}\right).</math>

Now consider if <math>y_0 = 0</math>, then <math>x_0 = \pm2\sqrt{2}.</math> The slope of the tangent line DNE. It follows that the tangent line is a vertical line, either <math>x = 2\sqrt{2}</math> or <math>x = -2\sqrt{2}</math>. Neither of them passes through the point <math>(0, -6).</math>

Therefore, the points on the curve that satisfy the requirements are <math>\color{blue}\left(\frac{8}{3}, -\frac{2}{3}\right),~\left(-\frac{8}{3}, -\frac{2}{3}\right)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 04/Solution 1

2020-04-29T13:27:22Z

DANIELDIBENEDETTO:

Let <math> (x_0, y_0) </math> be a point on the curve. Then this point satisfies

<math> x_0^2 + 2y_0^2 = 8.~~~~(1) </math>

To find the slope of the tangent line passing through <math> (x_0, y_0) </math>, use implicit differentiation. Differentiate both sides of the equation of the curve with respect to <math>x</math>. By the chain rule,

<math> 2x + 4yy' = 0. </math>

Let <math>y_0'</math> be the slope of the tangent line at point <math>(x_0, y_0)</math>. Assume <math>y_0 \neq 0</math>. Then

<math>y'_0 = -\frac{x_0}{2y_0}.</math>

Using the point-slope formula, the tangent line equation to the curve at the point <math>(x_0, y_0)</math> can be written as

<math> y - y_0 = -\frac{x_0}{2y_0}(x - x_0). </math>

Let this tangent line pass through <math>(0, -6)</math>. Putting <math>x = 0</math> and <math>y = -6</math>,

<math> -6 - y_0 = \frac{x_0^2}{2y_0}. </math>

Together with Equation (1) we can solve for the point. <math>(x_0, y_0) = \left(\frac{8}{3}, -\frac{2}{3}\right)</math> or <math>\left(-\frac{8}{3}, -\frac{2}{3}\right).</math>

Now consider if <math>y_0 = 0</math>, then <math>x_0 = \pm2\sqrt{2}.</math> The slope of the tangent line DNE. It follows that the tangent line is a vertical line, either <math>x = 2\sqrt{2}</math> or <math>x = -2\sqrt{2}</math>. Neither of them passes through the point <math>(0, -6).</math>

Therefore, the points on the curve that satisfy the requirements are <math>\left(\frac{8}{3}, -\frac{2}{3}\right),~\left(-\frac{8}{3}, -\frac{2}{3}\right)</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 03/Solution 1

2020-04-29T13:24:18Z

DANIELDIBENEDETTO:

We denote the limit by <math>\displaystyle L</math> and next we compute

<math>\displaystyle \log(L)=\lim_{x\to 0^+} \sin(x)\cdot \log(x)=\lim_{x\to 0^+}\frac{\log(x)}{\frac{1}{\sin(x)}}. </math>

We see that represented as a quotient above, then the limit for <math>\displaystyle\log(L)</math> is of the form <math>\displaystyle -\infty/\infty</math> and so, we apply L'Hôpital's Rule:

<math>\displaystyle\log(L)=\lim_{x\to 0^+}\frac{\frac{1}{x}}{\frac{-\cos(x)}{\sin^2(x)}}=\lim_{x\to 0^+}\frac{\sin^2(x)}{-x\cos(x)}.</math>

This limit is of the form <math>\displaystyle0/0</math>, so we apply again L'Hôpital's Rule and obtain:

<math>\displaystyle\log(L)=\lim_{x\to 0^+}\frac{2\sin(x)\cos(x)}{-\cos(x)-x\cdot (-\sin(x))}=0.</math>

Since <math>\displaystyle\log(L)=0</math>, we conclude that <math>\color{blue}\displaystyle L=1</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 03/Hint 2

2020-04-29T13:22:37Z

DANIELDIBENEDETTO:

You might need to use L'Hôpital's Rule, possibly more than once.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 03/Hint 2

2020-04-29T13:22:28Z

DANIELDIBENEDETTO:

You might need to use the L'Hôpital's Rule, possibly more than once.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (c)/Solution 1

2020-04-29T13:19:14Z

DANIELDIBENEDETTO:

We have

<math> \lim_{x\rightarrow 0^+}\frac{x}{|\tan x|}=\lim_{x\rightarrow 0^+}\frac{x}{\tan x}=1 ,</math>
but

<math>\lim_{x\rightarrow 0^-}\frac{x}{|\tan x|}=\lim_{x\rightarrow 0^-}\frac{x}{-\tan x}=-1, </math>
so <math>\color{blue}\text{the limit does not exist}</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (c)/Hint 1

2020-04-29T13:16:21Z

DANIELDIBENEDETTO:

Consider the limit as <math> x\rightarrow 0^+ </math> and as <math> x\rightarrow 0^- </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (c)/Hint 1

2020-04-29T13:16:02Z

DANIELDIBENEDETTO:

Consider both <math> x\rightarrow 0^+ </math> and <math> x\rightarrow 0^- </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (b)/Solution 1

2020-04-29T13:09:49Z

DANIELDIBENEDETTO:

For all <math>x</math>, we have <math> -1\le\sin(x)\le 1</math>, and therefore <math> -e^x\le e^x\sin(x)\le e^x</math>. Since
<math> \lim_{x\rightarrow -\infty} -e^x =\lim_{x\rightarrow -\infty} e^x =0,</math>
the squeeze theorem then gives <math>\color{blue} \lim_{x\to -\infty}e^x\sin(x)=0</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (b)/Hint 1

2020-04-29T13:05:00Z

DANIELDIBENEDETTO:

The Squeeze Theorem can help you here.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (a)/Hint 2

2020-04-29T13:00:29Z

DANIELDIBENEDETTO:

After simplifying the expression, divide by the highest power of <math>x</math> to see which terms will go to zero as <math>x\to-\infty</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (a)/Hint 2

2020-04-29T13:00:17Z

DANIELDIBENEDETTO: Created page with "After simplifying the expression, divide by the highest power of <math>x</math> to see which terms will go to zero as <math>x\to\infty</math>."

After simplifying the expression, divide by the highest power of <math>x</math> to see which terms will go to zero as <math>x\to\infty</math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (a)/Solution 1

2020-04-29T12:55:35Z

DANIELDIBENEDETTO:

Following the hint, we have

<math>\begin{align}
&\lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\\
&= \lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\cdot \frac{\sqrt{x^4+x^3+2}+x^2}{\sqrt{x^4+x^3+2}+x^2}\\
&= \lim_{x\to -\infty} \frac{x^4+x^3+2-x^4}{(5x+3)\cdot (\sqrt{x^4+x^3+2}+x^2)}\\
&= \lim_{x\to -\infty} \frac{x^3+2}{(5x+3)(\sqrt{x^4+x^3+2}+x^2)}.
\end{align}</math>

Now if we divide both the denominator and the numerator by <math>x^3</math>, we get

<math>\begin{align}
&\lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\\
&= \lim_{x\to -\infty} \frac{\frac{x^3+2}{x^3}}{\frac{5x+3}{x}\cdot \frac{\sqrt{x^4+x^3+2}+x^2}{x^2}}\\
&= \lim_{x\to -\infty} \frac{1+\frac{2}{x^3}}{\left(5+\frac{3}{x}\right)\cdot \left(\sqrt{\frac{x^4+x^3+2}{x^4}}+\frac{x^2}{x^2}\right)}\\
&= \lim_{x\to -\infty} \frac{1+\frac{2}{x^3}}{\left(5+\frac{3}{x}\right) \cdot \left(\sqrt{1+\frac{1}{x}+\frac{2}{x^4}} + 1\right)}\\
&= \frac{1}{5\cdot (1+1)}\\
&=\frac{1}{10},
\end{align}</math>

so <math>\color{blue} \lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3} = \frac{1}{10} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 02 (a)/Solution 1

2020-04-29T12:55:14Z

DANIELDIBENEDETTO:

<math>\lim_{x-\rightarrow -\infty}\frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}=\lim_{x-\rightarrow -\infty}\frac{(\sqrt{x^4+x^3+2}-x^2)(\sqrt{x^4+x^3+2}+x^2)}{(5x+3)(\sqrt{x^4+x^3+2}+x^2)}=\lim_{x-\rightarrow -\infty}\frac{x^3+2}{(5x+3)(\sqrt{x^4+x^3+2}+x^2)} =\lim_{x-\rightarrow -\infty}\frac{1+\frac{2}{x^3}}{(5+\frac{3}{x})(\sqrt{1+\frac{1}{x}+\frac{2}{x^4}}+1)}=\frac{1}{10} </math>

Following the hint, we have

<math>\begin{align}
&\lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\\
&= \lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\cdot \frac{\sqrt{x^4+x^3+2}+x^2}{\sqrt{x^4+x^3+2}+x^2}\\
&= \lim_{x\to -\infty} \frac{x^4+x^3+2-x^4}{(5x+3)\cdot (\sqrt{x^4+x^3+2}+x^2)}\\
&= \lim_{x\to -\infty} \frac{x^3+2}{(5x+3)(\sqrt{x^4+x^3+2}+x^2)}.
\end{align}</math>

Now if we divide both the denominator and the numerator by <math>x^3</math>, we get

<math>\begin{align}
&\lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3}\\
&= \lim_{x\to -\infty} \frac{\frac{x^3+2}{x^3}}{\frac{5x+3}{x}\cdot \frac{\sqrt{x^4+x^3+2}+x^2}{x^2}}\\
&= \lim_{x\to -\infty} \frac{1+\frac{2}{x^3}}{\left(5+\frac{3}{x}\right)\cdot \left(\sqrt{\frac{x^4+x^3+2}{x^4}}+\frac{x^2}{x^2}\right)}\\
&= \lim_{x\to -\infty} \frac{1+\frac{2}{x^3}}{\left(5+\frac{3}{x}\right) \cdot \left(\sqrt{1+\frac{1}{x}+\frac{2}{x^4}} + 1\right)}\\
&= \frac{1}{5\cdot (1+1)}\\
&=\frac{1}{10},
\end{align}</math>
so <math>\color{blue} \lim_{x\to -\infty} \frac{\sqrt{x^4+x^3+2}-x^2}{5x+3} = \frac{1}{10} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (f)/Hint 2

2020-04-29T10:58:41Z

DANIELDIBENEDETTO:

Choose a point at which you now the value of <math>f(x)</math> as the centre of the Taylor polynomial. Can you find a perfect cube close to <math>26</math>?

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (f)/Hint 2

2020-04-29T10:58:26Z

DANIELDIBENEDETTO:

Choose a point at which you now the value of <math>f(x)</math> as the centre of the Taylor polynomial. Is there a perfect cube close to <math>26</math>?

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (f)/Hint 2

2020-04-29T10:58:07Z

DANIELDIBENEDETTO: Created page with "Choose a point at which you now the value of <math>f(x)</math> as the centre of the Taylor polynomial. Is there a perfect cube close to 26?"

Choose a point at which you now the value of <math>f(x)</math> as the centre of the Taylor polynomial. Is there a perfect cube close to 26?

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (f)/Solution 1

2020-04-29T10:54:12Z

DANIELDIBENEDETTO:

Let <math> f(x)=x^{\frac{1}{3}} </math>, then <math> f'(x)=\frac{1}{3}x^{-\frac{2}{3}}</math>. Since we know that <math> f(27)=3 </math>, this is where we will centre the Taylor polynomial. Then we have

<math> T_1(26)=f(27)+f'(27)\cdot (26-27) = 3 + \frac{1}{27}\cdot (-1)=\frac{80}{27}.</math>

We conclude that
<math>\color{blue} \sqrt[3]{26} \approx \frac{80}{27} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (f)/Hint 1

2020-04-29T10:47:40Z

DANIELDIBENEDETTO:

Let <math> f(x)=x^{\frac{1}{3}} </math>, and approximate <math> f(26) </math> using the first Taylor polynomial centred at a nearby point.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (e)/Solution 1

2020-04-29T10:46:10Z

DANIELDIBENEDETTO:

Taking the logaritm of the orignal function, we have <math> \log f=x\log (2+\cos x) </math>,
Computing its derivative (using the product rule for the right hand side) gives
<math> \frac{f'(x)}{f(x)}=\log(2+\cos x)+\frac{x}{2+\cos x } (-\sin x) </math>
so after rearranging, we obtain
<math>\color{blue} f'(x) = (2+\cos (x))^x \left( \log \left( 2 + \cos(x) \right) - \frac{x \sin (x) }{ 2 + \cos(x)} \right) </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (e)/Solution 1

2020-04-29T10:45:26Z

DANIELDIBENEDETTO:

Taking the logaritm of the orignal function, we have <math> \log f=x\log (2+\cos x) </math>,
Computing its derivative (using the product rule for the right hand side) gives
<math> \frac{f'(x)}{f(x)}=\log(2+\cos x)+\frac{x}{2+\cos x } (-\sin x) </math>
so we obtain
<math> f'(x)=(2+\cos x)^x\left[\ln(2+\cos x)-\frac{x\sin x}{2+\cos x}\right] </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (e)/Hint 2

2020-04-29T10:44:25Z

DANIELDIBENEDETTO: Created page with "Recall that <math> (\log(f(x))'= \frac{f'(x)}{f(x)} </math>."

Recall that <math> (\log(f(x))'= \frac{f'(x)}{f(x)} </math>.

Science:Math Exam Resources/Courses/MATH100/December 2018/Question 01 (e)/Hint 1

2020-04-29T10:42:53Z

DANIELDIBENEDETTO:

Take the logarithm of both sides and then compute the derivative.