Science:Math Exam Resources/Courses/MATH110/April 2019/Question 05 (c)/Solution 1

Recall that a function f(x) is concave down at all points where its second derivative is negative; that is, f ′′(x) < 0.

From part (a), we know that

$f^{'} (x) = \frac{x^{4} + 3 x^{2}}{(x^{2} + 1)^{2}} .$

Using the quotient rule again, we find that

$f^{″} (x) = \frac{(x^{4} + 3 x^{2})^{'} (x^{2} + 1)^{2} - (x^{4} + 3 x^{2}) ((x^{2} + 1)^{2})^{'}}{(x^{2} + 1)^{4}} = \frac{(4 x^{3} + 6 x) (x^{2} + 1)^{2} - (x^{4} + 3 x^{2}) (2 (x^{2} + 1) \cdot 2 x)}{(x^{2} + 1)^{4}} .$

This simplifies to

$f^{″} (x) = \frac{(x^{2} + 1) \cdot (x (4 x^{2} + 6) (x^{2} + 1) - 4 x (x^{4} + 3 x^{2}))}{(x^{2} + 1)^{4}} = \frac{x (6 - 2 x^{2})}{(x^{2} + 1)^{3}} = - \frac{2 x (x^{2} - 3)}{(x^{2} + 1)^{3}} .$

The denominator is always positive (by the same reasoning as in the solution to part (b)), so f "(x) will be negative whenever the numerator is negative, in symbols,

$f^{″} (x) < 0 whenever x (x^{2} - 3) < 0 .$

To figure out for which values of x the inequality $x (x^{2} - 3) < 0$ holds, factor the expression as

$x (x^{2} - 3) = x (x - \sqrt{3}) (x + \sqrt{3}) .$

We do this because it's easier to figure out for what x each of $x$ , $x - \sqrt{3}$ or $x + \sqrt{3}$ is negative and we know that a product of three numbers is negative if one of them is negative and the other two are positive or if all three are negative. Therefore $x (x - \sqrt{3}) (x + \sqrt{3}) < 0$ if $x < - \sqrt{3}$ or if $x$ is in the interval $(0, \sqrt{3})$ , in which case $x - \sqrt{3} < 0$ and $x (x + \sqrt{3})$ is positive.