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.

$f(x)=e^{2x}+e^{-x}$

find each local minimum, if any. select the correct choice below and, if necessary, fill in the answer boxes
to complete your choice.

(type exact answers.)

○ a. the function has a local minimum value at one value of x. the minimum value is $f(\square)=\square$

○ b. the function has a local minimum value at three values of x. in increasing order of x - value, the
minimum values are $f(\square)=\square,f(\square)=\square$, and $f(\square)=\square$

○ c. the function has a local minimum value at two values of x. in increasing order of x - value, the
minimum values are $f(\square)=\square$ and $f(\square)=\square$

○ d. there are no local minima.

Explanation:

Step1: Find the derivative of the function

Using the chain - rule, if \(y = e^{2x}+e^{-x}\), then \(y^\prime=f^\prime(x)=2e^{2x}-e^{-x}\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(2e^{2x}-e^{-x}=0\). Multiply through by \(e^{x}\) (since \(e^{x}
eq0\) for all real \(x\)) to get \(2e^{3x}-1 = 0\). Then \(e^{3x}=\frac{1}{2}\), and \(3x=\ln(\frac{1}{2})=-\ln 2\), so \(x =-\frac{\ln 2}{3}\).

Step3: Determine the intervals of increase and decrease

• Choose a test point in the interval \((-\infty,-\frac{\ln 2}{3})\), say \(x=-1\). Then \(f^\prime(-1)=2e^{-2}-e^{1}=\frac{2}{e^{2}}-e\lt0\).
• Choose a test point in the interval \((-\frac{\ln 2}{3},\infty)\), say \(x = 0\). Then \(f^\prime(0)=2e^{0}-e^{0}=2 - 1=1\gt0\).

So the function \(y = f(x)\) is decreasing on the interval \((-\infty,-\frac{\ln 2}{3})\) and increasing on the interval \((-\frac{\ln 2}{3},\infty)\).

Step4: Find the local extreme values

Since the function changes from decreasing to increasing at \(x =-\frac{\ln 2}{3}\), by the first - derivative test, the function has a local minimum at \(x =-\frac{\ln 2}{3}\).
Calculate \(f(-\frac{\ln 2}{3})=e^{2(-\frac{\ln 2}{3})}+e^{-(-\frac{\ln 2}{3})}=e^{-\frac{2\ln 2}{3}}+e^{\frac{\ln 2}{3}}=\frac{1}{2^{\frac{2}{3}}}+2^{\frac{1}{3}}=\frac{1 + 2}{2^{\frac{2}{3}}}=\frac{3}{2^{\frac{2}{3}}}\)

Answer:

• a. The function \(f(x)\) is decreasing on the interval \((-\infty,-\frac{\ln 2}{3})\) and increasing on the interval \((-\frac{\ln 2}{3},\infty)\).
• b. The function has a local minimum value. The minimum value is \(f(-\frac{\ln 2}{3})=\frac{3}{2^{\frac{2}{3}}}\), so the answer is A. The function has a local minimum value at one value of \(x\). The minimum value is \(f(-\frac{\ln 2}{3})=\frac{3}{2^{\frac{2}{3}}}\)