Question

question 6
a. ( f ( - x ) = f ( 2 x ) )
b. ( f ( - x ) = - f ( 2 x ) )
c. ( f ( - x ) = - f ( x ) )
d. ( f ( - x ) = f ( x ) )
show that ( f ( x ) = cosh x ) is an even function.
( cosh ( - x ) = \frac { 1 } { 2 } ( e ^ { - ( - x ) } + e ^ { - x } ) )
( = \frac { 1 } { 2 } ( e ^ { - x } + e ^ { x } ) )
(simplify your answer.)
=

• cosh ( 2 x )

( cosh x )

• cosh x

( cosh ( 2 x ) )

Explanation:

Step1: Substitute \(-x\) into \(\cosh x\)

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

Step2: Simplify the expression

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

Answer:

Since \(\cosh(-x)=\frac{1}{2}(e^{-x}+e^{x})\) and \(\cosh x=\frac{1}{2}(e^{x}+e^{-x})\), so \(\cosh(-x)=\cosh x\). By the definition of an even - function \(f(-x)=f(x)\), the function \(y = \cosh x\) is an even function.