find the critical points of the following function.\n\n$f(x)=\\frac{e^{x}+e^{-x}}{11}$\n\nselect the correct…

find the critical points of the following function.\n\n$f(x)=\\frac{e^{x}+e^{-x}}{11}$\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the critical point(s) occur(s) at $x=$\n(use a comma to separate answers as needed.)\nb. there are no critical points.

find the critical points of the following function.\n\n$f(x)=\\frac{e^{x}+e^{-x}}{11}$\n\nselect the correct choice below and, if necessary, fill in the answer box to complete your choice.\n\na. the critical point(s) occur(s) at $x=$\n(use a comma to separate answers as needed.)\nb. there are no critical points.

Answer

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\neq0 )): ( 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 )