Question

find the critical points of the following function.

$f(x)=\frac{e^{x}+e^{-x}}{11}$

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the critical point(s) occur(s) at $x=$
(use a comma to separate answers as needed.)
b. there are no critical points.

Explanation:

Step1: Find the derivative of \( f(x) \)

Using the sum - rule and the constant - multiple rule of differentiation. If \( f(x)=\frac{e^{x}+e^{-x}}{11}=\frac{1}{11}e^{x}+\frac{1}{11}e^{-x} \), then \( f^{\prime}(x)=\frac{1}{11}e^{x}-\frac{1}{11}e^{-x} \) (since \( \frac{d}{dx}(e^{x}) = e^{x} \) and \( \frac{d}{dx}(e^{-x})=-e^{-x} \)).

Step2: Set \( f^{\prime}(x) = 0 \) and solve for \( x \)

Set \( \frac{1}{11}e^{x}-\frac{1}{11}e^{-x}=0 \). Multiply through by \( 11 \) to get \( e^{x}-e^{-x}=0 \). Rewrite \( e^{-x}=\frac{1}{e^{x}} \), so the equation becomes \( e^{x}-\frac{1}{e^{x}} = 0 \). Let \( t = e^{x}(t>0) \), then \( t-\frac{1}{t}=0 \). Multiply through by \( t \) (since \( t
eq0 \)): \( t^{2}-1 = 0 \), which factors as \( (t - 1)(t + 1)=0 \). Since \( t=e^{x}>0 \), \( t = 1 \). If \( e^{x}=1 \), then \( x = 0 \) (because \( y = e^{x} \) and \( e^{0}=1 \)).

Answer:

A. The critical point(s) occur(s) at \( x = 0 \)