Question

consider the equation below. (if an answer does not exist, enter dne.)

$f(x)=e^{9x}+e^{-x}$

(a) find the interval on which $f$ is increasing. (enter your answer using interval notation.)

find the interval on which $f$ is decreasing. (enter your answer using interval notation.)

(b) find the local minimum and maximum values of $f$.

local minimum value

local maximum value

(c) find the inflection point.

$(x,y)=( )$

find the interval on which $f$ is concave up. (enter your answer using interval notation.)

find the interval on which $f$ is concave down. (enter your answer using interval notation.)

Explanation:

Step1: Find the first derivative

Using the chain rule, if \(y = e^{u}\), then \(y^\prime=e^{u}\cdot u^\prime\).
For \(f(x)=e^{9x}+e^{-x}\), \(f^\prime(x)=9e^{9x}-e^{-x}=\frac{9e^{10x} - 1}{e^{x}}\)

Step2: Find critical points

Set \(f^\prime(x) = 0\), so \(9e^{10x}-1 = 0\).
\(e^{10x}=\frac{1}{9}\), then \(10x=\ln(\frac{1}{9})=-\ln(9)\), \(x =-\frac{\ln(9)}{10}=-\frac{\ln(3^{2})}{10}=-\frac{\ln(3)}{5}\)

Step3: Determine increasing and decreasing intervals

• Test an interval to the left of \(x =-\frac{\ln(3)}{5}\), say \(x=-1\). \(f^\prime(-1)=9e^{-9}-e^{1}<0\)
• Test an interval to the right of \(x =-\frac{\ln(3)}{5}\), say \(x = 0\). \(f^\prime(0)=9 - 1=8>0\)

So \(f(x)\) is decreasing on \((-\infty,-\frac{\ln(3)}{5})\) and increasing on \((-\frac{\ln(3)}{5},\infty)\)

Step4: Find local minimum and maximum

Since \(f(x)\) changes from decreasing to increasing at \(x =-\frac{\ln(3)}{5}\), the local minimum value is \(f(-\frac{\ln(3)}{5})=e^{9(-\frac{\ln(3)}{5})}+e^{-(-\frac{\ln(3)}{5})}=e^{-\frac{9\ln(3)}{5}}+e^{\frac{\ln(3)}{5}}=\frac{1}{3^{\frac{9}{5}}}+3^{\frac{1}{5}}=\frac{1 + 3^{2}}{3^{\frac{9}{5}}}=\frac{10}{3^{\frac{9}{5}}}\) and there is no local maximum (because the function only has one critical point and the function changes from decreasing to increasing)

Step5: Find the second derivative

\(f^\prime(x)=9e^{9x}-e^{-x}\), then \(f^{\prime\prime}(x)=81e^{9x}+e^{-x}=\frac{81e^{10x}+1}{e^{x}}\)
Since \(e^{10x}>0\) for all \(x\in R\), \(f^{\prime\prime}(x)>0\) for all \(x\in R\)

Answer:

(a) Increasing: \((-\frac{\ln(3)}{5},\infty)\), Decreasing: \((-\infty,-\frac{\ln(3)}{5})\)
(b) Local minimum value: \(\frac{10}{3^{\frac{9}{5}}}\), Local maximum value: DNE
(c) Inflection point: DNE, Concave up: \((-\infty,\infty)\), Concave down: DNE