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

MATH104 December 2011

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Question 02 (c)
Consider the function $f(x)={\frac {x^{3}+x^{2}-2x-3}{x^{2}-3}}.$ Its first and second derivatives are given by $f'(x)={\frac {(x^{2}-1)(x^{2}-6)}{(x^{2}-3)^{2}}},\qquad f''(x)={\frac {2x(x^{2}+9)}{(x^{2}-3)^{3}}}.$ On what intervals is f(x) increasing? On what intervals is f(x) decreasing?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
A differentiable function is increasing on an interval if its derivative is positive on that interval. What would a sign table of the derivative tell you then?

Hint 2
If you don't like a sign table: In question 2 (a), we determined where f'(x)=0 or does not exist, i.e. the critical points of f(x). Try testing the sign of f' for values between two adjacent critical points.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Factoring the derivative yields $f'(x)={\frac {(x-1)(x+1)(x-{\sqrt {6}})(x+{\sqrt {6}})}{(x-{\sqrt {3}})^{2}(x+{\sqrt {3}})^{2}}}$ Now the following sign table is straight forward: $(-\infty ;-{\sqrt {6}})$ $(-{\sqrt {6}};-{\sqrt {3}})$ $(-{\sqrt {3}};-1)$ $\displaystyle (-1;1)$ $(1;{\sqrt {3}})$ $({\sqrt {3}};{\sqrt {6}})$ $({\sqrt {6}};\infty )$ $\displaystyle x-1$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle x+1$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $x-{\sqrt {6}}$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $x+{\sqrt {6}}$ $\displaystyle -$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $(x-{\sqrt {3}})^{2}$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $(x+{\sqrt {3}})^{2}$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle +$ $\displaystyle f'(x)$ $\displaystyle +$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $\displaystyle -$ $\displaystyle -$ $\displaystyle +$ $\displaystyle f(x)$ $\displaystyle \nearrow$ $\displaystyle \searrow$ $\displaystyle \searrow$ $\displaystyle \nearrow$ $\displaystyle \searrow$ $\displaystyle \searrow$ $\displaystyle \nearrow$ And so we can conclude that the function is increasing on: $(-\infty ,-{\sqrt {6}})\cup (-1,1)\cup ({\sqrt {6}},\infty )$ and decreasing on: $(-{\sqrt {6}},-{\sqrt {3}})\cup (-{\sqrt {3}},-1)\cup (1,{\sqrt {3}})\cup ({\sqrt {3}},{\sqrt {6}})$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. If you don't like a sign table: From Question 2 (a), we know that the critical points of ƒ(x), when ƒ'(x) or does not exist, are at x = ±1, ±√6, and ±√3. Testing a value less than -√6, such as -3, we get ƒ'(-3) = 2/3 > 0. Testing a point between -√6 and -√3, such as -2, yields ƒ'(-2) = -2 < 0. Testing a point between -√3 and -1, such as -1.5, we get ƒ'(-1.5) = -25/3 < 0. Finally, we test a point between -1 and 1, such as 0, and get ƒ'(0) = 2/3 > 0. Normally, we would have to test points between the positive critical points as well, but since our function only has x2 terms in it (i.e. ƒ' is even), 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 ƒ(x) < 0 between -√6 and -√3 as well as between √3 and √6.) Therefore: ƒ(x) is increasing on the intervals $(-\infty ,-{\sqrt {6}})\cup (-1,1)\cup ({\sqrt {6}},\infty )$ and decreasing on the intervals $(-{\sqrt {6}},-{\sqrt {3}})\cup (-{\sqrt {3}},-1)\cup (1,{\sqrt {3}})\cup ({\sqrt {3}},{\sqrt {6}})$. Note that the critical points are not included in the intervals of increase and decrease.

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