# How do you use the second derivative test to find where the function f(x) = (5e^x)/(5e^x + 6) is concave up, concave down, and inflection points?

Aug 3, 2015

$f \left(x\right)$ will be concave down on $\left(\ln \left(1.2\right) , + \infty\right)$ and concave up on $\left(- \infty , \ln \left(1.2\right)\right)$.

#### Explanation:

The second derivative test allows you to determine the intervals on which a function is concave up or concave down by examining the sign of the second derivative around the inflexion points.

Inflexion points are points for which ${f}^{' '} \left(x\right) = 0$.

So, start by calculating the first derivative of $f \left(x\right)$ - use the quotient rule

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(5 {e}^{x}\right)\right] \cdot \left(5 {e}^{x} + 6\right) - 5 {e}^{x} \cdot \frac{d}{\mathrm{dx}} \left(5 {e}^{x} + 6\right)}{5 {e}^{x} + 6} ^ 2$

${f}^{'} = \frac{5 {e}^{x} \cdot \left(5 {e}^{x} + 6\right) - 5 {e}^{x} \cdot 5 {e}^{x}}{5 {e}^{x} + 6} ^ 2$

${f}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{5 {e}^{x}}}} + 30 {e}^{x} - \textcolor{red}{\cancel{\textcolor{b l a c k}{5 {e}^{x}}}}}{5 {e}^{x} + 6} ^ 2 = \frac{30 {e}^{x}}{5 {e}^{x} + 6} ^ 2$

Next, calculate the second derivative by using the quotient and chain rules

$\frac{d}{\mathrm{dx}} \left({f}^{'} \left(x\right)\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left(30 {e}^{x}\right)\right] \cdot {\left(5 {e}^{x} + 6\right)}^{2} - 30 {e}^{x} \cdot \frac{d}{\mathrm{dx}} {\left(5 {e}^{x} + 6\right)}^{2}}{{\left(5 {e}^{x} + 6\right)}^{2}} ^ 2$

${f}^{' '} = \frac{30 {e}^{x} \cdot {\left(5 {e}^{x} + 6\right)}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} - 30 {e}^{x} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(5 {e}^{x} + 6\right)}}} \cdot 5 {e}^{x}}{5 {e}^{x} + 6} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}$

${f}^{' '} = \frac{150 {e}^{2 x} + 180 {e}^{x} - 300 {e}^{2 x}}{5 {e}^{x} + 6} ^ 3 = - \frac{30 {e}^{x} \left(5 {e}^{x} - 6\right)}{5 {e}^{x} + 6} ^ 3$

FInd the critical poin(s) of the function by calculating

${f}^{' '} = 0$

$- \frac{30 {e}^{x} \left(5 {e}^{x} - 6\right)}{5 {e}^{x} + 6} ^ 3 = 0 \iff \left(5 {e}^{x} - 6\right) = 0$

This will get you

$5 {e}^{x} = 6$

${e}^{x} = \frac{6}{5} = 1.2$

Take the natual log of both sides of the equation to get

$\ln \left({e}^{x}\right) = \ln \left(1.2\right) \implies x \cdot \ln e = \ln \left(1.2\right) \implies x = \textcolor{g r e e n}{\ln \left(1.2\right)}$

Now investigate the sign of the sign derivative for values smaller than $\ln \left(1.2\right)$ and larger than $\ln \left(1.2\right)$.

SInce ${e}^{a} > 0$ for any real number $a$, the sign of the second derivative will depend on the numerator of the fraction, $- 30 {e}^{x} \left(5 {e}^{x} - 6\right)$.

So, the two intervals that you're going to look at are

• $\left(- \infty , \ln \left(1.2\right)\right)$

For values smaller than $\ln \left(1.2\right)$, $\left(5 {e}^{x} - 6\right)$ will be negative, which means that ${f}^{' '}$ will be positive.

As a result, $f \left(x\right)$ will be concave up on this interval.

• $\left(\ln \left(1.2\right) , + \infty\right)$

This time, $\left(5 {e}^{x} - 6\right) > 0$, so it follows that ${f}^{' '}$ will be negative.

This implies that $f \left(x\right)$ is concave down on this interval.

The point $\left(\ln \left(1.2\right) , f \left(\ln \left(1.2\right)\right)\right)$ will be the only Inflexion point for this function.