# How do you use the first and second derivatives to sketch f(x)=x^2/(x^2+5)?

Feb 5, 2017

See the explanation below

graph{x^2/(x^2+5) [-5, 5, -2.5, 2.5]}

#### Explanation:

First we note that, as ${x}^{2} + 5 > 0$ for all $x$, the function is defined and continuous in all of $\mathbb{R}$.

We also note that as the numerator and denominator only contain terms of even degree in $x$ the function is even, that is:

$f \left(- x\right) = f \left(x\right)$

so the graph will be symmetric with respect to the $y$ axis.

We then evaluate the behavior at the limits of the domain:

${\lim}_{x \to - \infty} {x}^{2} / \left({x}^{2} + 5\right) = {\lim}_{x \to + \infty} {x}^{2} / \left({x}^{2} + 5\right) = 1$

so on both sides the line $y = 1$ is an asymptote.

Now we calculate the first derivative using the quotient rule:

$f ' \left(x\right) = \frac{2 x \left({x}^{2} + 5\right) - 2 {x}^{3}}{{x}^{2} + 5} ^ 2 = \frac{10 x}{{x}^{2} + 5} ^ 2$

The only critical point is then $x = 0$ and we have:

$\left\{\begin{matrix}f ' \left(x\right) < 0 \text{ for " x < 0 \\ f'(x) > 0 " for } x > 0\end{matrix}\right.$

so $f \left(x\right)$ is decreasing in $\left(- \infty , 0\right)$ and increasing in $\left(0 , + \infty\right)$, which means the value for $x = 0$ is a local minimum.

As the value of the minimum is $f \left(0\right) = 0$ and the function is positive everywhere else, this is also an absolute minimum.

Calculating the second derivative:

$f ' ' \left(x\right) = \frac{10 {\left({x}^{2} + 5\right)}^{2} - 40 {x}^{2} \left({x}^{2} + 5\right)}{{x}^{2} + 5} ^ 4 = \frac{10 {x}^{2} + 50 - 40 {x}^{2}}{{x}^{2} + 5} ^ 3 = - 10 \frac{3 {x}^{2} - 5}{{x}^{2} + 5} ^ 3$

The points of inflection are then the roots of:

$3 {x}^{2} - 5 = 0$

$x = \pm \sqrt{\frac{5}{3}}$

to solve the inequation $f ' ' \left(x\right) > 0$ we can then note that:

(i) the denominator is always positive

(ii) the numerator $3 {x}^{2} - 5$ is a second order polynomial with two roots, so it is negative in the interval between the roots and positive outside.

(iii) the sign of $f ' ' \left(x\right)$ is the opposite of the numerator's.

Then:

$f \left(x\right)$ is concave down for $x \in \left(- \infty , - \sqrt{\frac{5}{3}}\right)$

$f \left(x\right)$ is concave up for $x \in \left(- \sqrt{\frac{5}{3}} , \sqrt{\frac{5}{3}}\right)$

$f \left(x\right)$ is concave down for $x \in \left(\sqrt{\frac{5}{3}} , + \infty\right)$

Mar 29, 2017