Question

a. find the open intervals on which the function is increasing and those on which it is decreasing
b. identify the functions local extreme values, if any, saying where they occur.
f(x)=e^{6x}+e^{-x}
a. on what open intervals, if any, is the function increasing? select the correct choice below and, if necessary,
fill in the answer box to complete your choice.
a. the function f is increasing on the open interval(s)
(simplify your answer. type your answer in interval notation. type exact an answer use a comma to
separate answers as needed.)
b. the function is never increasing.

Explanation:

Step1: Find the derivative of the function

The derivative of \(y = e^{6x}+e^{-x}\) using the chain rule \((e^{u})^\prime=e^{u}\cdot u^\prime\).
For \(y = e^{6x}\), let \(u = 6x\), then \(y^\prime=e^{6x}\cdot6\). For \(y = e^{-x}\), let \(u=-x\), then \(y^\prime=e^{-x}\cdot(- 1)\).
So \(f^\prime(x)=6e^{6x}-e^{-x}=\frac{6e^{7x}-1}{e^{x}}\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), i.e., \(\frac{6e^{7x}-1}{e^{x}}=0\). Since \(e^{x}>0\) for all \(x\in R\), we solve \(6e^{7x}-1 = 0\).
\(e^{7x}=\frac{1}{6}\), then \(7x=\ln(\frac{1}{6})=-\ln6\), and \(x =-\frac{\ln6}{7}\).

Step3: Determine the sign of the derivative

Choose a test - point in the interval \((-\infty,-\frac{\ln6}{7})\), say \(x = - 1\).
\(f^\prime(-1)=6e^{-6}-e^{1}=\frac{6}{e^{6}}-e<0\).
Choose a test - point in the interval \((-\frac{\ln6}{7},\infty)\), say \(x = 0\).
\(f^\prime(0)=6e^{0}-e^{0}=6 - 1=5>0\).

Answer:

A. The function \(f\) is increasing on the open interval\((-\frac{\ln6}{7},\infty)\)