Science:Math Exam Resources/Courses/MATH102/December 2012/Question B 02

MATH102 December 2012

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Question B 02

Consider the function $f(x)={\frac {x}{x^{2}+1}}$ defined on the whole real line. The zeros of $\ f'$, increasing order, are c1 = ___ and c2 = ___. The zeros of $\ f''$, in increasing order, are r1 = ___, r2 = ___ and r3 = ___. In each empty cell of the tables below, enter a + or - to indicate sign of $\ f$, $\ f'$ and $\ f''$ as appropriate. $(-\infty ,0)$ 0 $(0,\infty )$ $\ f(x)$ 0 $(-\infty ,c_{1})$ c1 $\displaystyle (c_{1},c_{2})$ c2 $(c_{2},\infty )$ $\ f'(x)$ 0 0 $(-\infty ,r_{1})$ r1 $\displaystyle (r_{1},r_{2})$ r2 $\displaystyle (r_{2},r_{3})$ r3 $(r_{3},\infty )$ $\ f''(x)$ 0 0 0

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
Calculate the first and second derivatives of $f(x)$ and set them equal to zero to find the critical points.

Hint 2
To fill in the rest of the tables, for each cell choose a point in the appropriate interval (which will partially be determined by the critical points you've already found). Plug it into the appropriate function/derivative to determine whether it is positive or negative. Save yourself some work - you don't need to find the exact value, just whether it is positive or negative.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Before taking derivatives, we will fill in the first table, by testing values in the two intervals listed. Note that $x^{2}+1$ will always be positive, so when calculating whether the function/derivatives are positive or negative, it is sufficient to look at the numerator as the denominator will always be positive. $(-\infty ,0)$, test point $x=-1$ $f(-1)={\frac {-1}{(-1)^{2}+1}}={\frac {-1}{2}}<0$ $(0,\infty )$, test point $x=1$ $f(1)={\frac {1}{(1)^{2}+1}}={\frac {1}{2}}>0$ So the first table should be filled in as: $(-\infty ,0)$ 0 $(0,\infty )$ ƒ(x) - 0 + Proceeding to the second table, we will calculate the first derivative using the quotient rule: $f'(x)={\frac {1*(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}}={\frac {1-x^{2}}{(x^{2}+1)^{2}}}$ Setting equal to zero and solving for x gives: $\displaystyle 0=1-x^{2}$ That is, $\displaystyle x=-1$, $\displaystyle x=1$. These are then our values for $c_{1}$ and $c_{2}$. We can now test points in the three intervals labeled in the table. $(-\infty ,-1)$, test point $x=-2$ $f'(-2)={\frac {1-(-2)^{2}}{((-2)^{2}+1)^{2}}}={\frac {-3}{25}}<0$ $(-1,1)$, test point $x=0$ $f'(0)={\frac {1-(0)^{2}}{((0)^{2}+1)^{2}}}={\frac {1}{1}}>0$ $(1,\infty )$, test point $x=2$ $f'(2)={\frac {1-(2)^{2}}{((2)^{2}+1)^{2}}}={\frac {-3}{25}}<0$ Filling in the table should look like this: $(-\infty ,c_{1})$ c1 $\displaystyle (c_{1},c_{2})$ c2 $(c_{2},\infty )$ ƒ'(x) - 0 + 0 - Finally, we compute the second derivative, again using the quotient rule: ${\begin{aligned}f''(x)&={\frac {-2x(x^{2}+1)^{2}-(1-x^{2})(2)(x^{2}+1)(2x)}{(x^{2}+1)^{4}}}\\&={\frac {-2x(x^{2}+1)-(1-x^{2})(4x)}{(x^{2}+1)^{3}}}\\&={\frac {-2x^{3}-2x-4x+4x^{3}}{(x^{2}+1)^{3}}}\\&={\frac {2x^{3}-6x}{(x^{2}+1)^{3}}}\\&={\frac {2x(x^{2}-3)}{(x^{2}+1)^{3}}}\end{aligned}}$ Again solving for where $\displaystyle f''(x)=0$ we have $\displaystyle 0=2x(x^{2}-3)$ which yields critical points of $x=-{\sqrt {3}},x=0$ and $x={\sqrt {3}}$, which will be $r_{1},r_{2},r_{3}$ respectively. When testing points in each of the four intervals of the table, it is worth noting that ${\sqrt {3}}$ is somewhere between 1 and 2. $(-\infty ,-{\sqrt {3}})$, test point $x=-2$ $f''(-2)={\frac {2(-2)((-2)^{2}-3)}{({-2}^{2}+1)^{3}}}={\frac {-4}{5^{3}}}<0$ $(-{\sqrt {3}},0)$, test point $x=-1$ $f''(-1)={\frac {2(-1)((-1)^{2}-3)}{({-1}^{2}+1)^{3}}}={\frac {4}{2^{3}}}>0$ $(0,{\sqrt {3}})$, test point $x=1$ $f''(1)={\frac {2(1)((1)^{2}-3)}{({1}^{2}+1)^{3}}}={\frac {-4}{2^{3}}}<0$ $({\sqrt {3}},\infty ,)$, test point $x=2$ $f''(2)={\frac {2(2)((2)^{2}-3)}{({2}^{2}+1)^{3}}}={\frac {4}{5^{3}}}>0$ Filling in the table yields: $(-\infty ,r_{1})$ r1 $\displaystyle (r_{1},r_{2})$ r2 $\displaystyle (r_{2},r_{3})$ r3 $(r_{3},\infty )$ ƒ(x) - 0 + 0 - 0 +

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