# Question 2237b

Jun 13, 2017

a) $f$ is strictly increasing on x in ]-ln7/8,oo[, and strictly decreasing on x in ]-oo,-ln7/8[.
b) $x = - \ln \frac{7}{8}$ is a local minimum.
c) There is no inflexion point.

#### Explanation:

$f \left(x\right) = {e}^{7 x} + {e}^{- x}$
As you have managed to find, $f ' \left(x\right) = 7 {e}^{7 x} - {e}^{- x}$

a) To find the interval on which $f$ is increasing or decreasing, we find the interval on which $f ' \left(x\right) > \mathmr{and} < 0$
$f ' \left(x\right) > 0$
$7 {e}^{7 x} > {e}^{- x}$
$7 {e}^{8 x} > 1$
$\ln \left({e}^{8 x}\right) > \ln \left(\frac{1}{7}\right)$
$8 x > - \ln 7$
$x > - \ln \frac{7}{8}$
Likewise,
$f ' \left(x\right) < 0$
$7 {e}^{7 x} < {e}^{- x}$
$7 {e}^{8 x} < 1$
$\ln \left({e}^{8 x}\right) < \ln \left(\frac{1}{7}\right)$
$8 x < - \ln 7$
$x < - \ln \frac{7}{8}$

Therefore, $f$ is strictly increasing on x in ]-ln7/8,oo[, and strictly decreasing on x in ]-oo,-ln7/8[#.

b) To find the maxima or minima (critical points), we find the $x$ such that $f ' = 0$ at those points.
$f ' \left(x\right) = 0$
$7 {e}^{7 x} = {e}^{- x}$
$7 {e}^{8 x} = 1$
$\ln \left({e}^{8 x}\right) = \ln \left(\frac{1}{7}\right)$
$8 x = - \ln 7$
$x = - \ln \frac{7}{8}$

Therefore, $x = - \ln \frac{7}{8}$ is a critical point. As we have determined in part (a) that $f$ is decreasing on $x < - \ln \frac{7}{8}$ and increasing on $x > - \ln \frac{7}{8}$, we can conclude that $x = - \ln \frac{7}{8}$ is a local minimum.

c) To find the inflexion point, at high school level or in precalculus, we find the $x$ such that $f ' ' = 0$ at those points.
$f ' \left(x\right) = 7 {e}^{7 x} - {e}^{- x}$
$f ' ' \left(x\right) = 49 {e}^{7 x} + {e}^{- x}$
$f ' ' \left(x\right) = 0$
$49 {e}^{7 x} + {e}^{- x} = 0$
${e}^{8 x} = - \frac{1}{49}$

As ${e}^{x} > 0$ for all $x \in R$, there is no solution for ${e}^{8 x} = - \frac{1}{49}$ Therefore, there is no inflexion point for $f$.