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

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

Using the quotient rule again, we find that

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

This simplifies to

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

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,

$f''(x)<0\,{\text{ whenever }}\,x(x^{2}-3)<0.$

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 $\color {blue}x<-{\sqrt {3}}$ or if $x$ is in the interval $\color {blue}(0,{\sqrt {3}})$, in which case $x-{\sqrt {3}}<0$ and $x(x+{\sqrt {3}})$ is positive.