Question

show that ( f(x)=cosh x ) is an even function
( cosh (-x)=\frac{1}{2}left(e^{-x}+e^{-(-x)}
ight) )
( =\frac{1}{2}left(e^{-x}+e^{x}
ight) )
( =cosh x )
(b) graph ( f(x)=cosh x ) using a graphing utility. choose the correct graph below
a.
( -6,6,1 ) by ( -6,7,1 )
b.
( -6,6,1 ) by ( -6,6,1 )
c.
( -6,6,1 ) by ( -1,6,1 )
d.
( -6,6,1 ) by ( -6,6,1 )

Explanation:

Step1: Recall the definition of an even function

An even function satisfies \(f(-x)=f(x)\) for all \(x\) in its domain.

Step2: Substitute \(-x\) into the function \(y = \cosh(x)\)

Given \(f(x)=\cosh(x)=\frac{1}{2}(e^{x}+e^{-x})\), then \(f(-x)=\frac{1}{2}(e^{-x}+e^{-(-x)})\).

Step3: Simplify \(f(-x)\)

Since \(e^{-(-x)} = e^{x}\), we have \(f(-x)=\frac{1}{2}(e^{-x}+e^{x})\).

Step4: Compare \(f(-x)\) and \(f(x)\)

We can see that \(f(-x)=\frac{1}{2}(e^{-x}+e^{x})=\frac{1}{2}(e^{x}+e^{-x}) = f(x)\).

Answer:

C. \(f(-x)=f(x)\)