Question

(b) graph ( f(x)=cosh x ) using a graphing utility. choose the correct graph below.
(c) to show that, for every ( x,(cosh x)^{2}-(sinh x)^{2}=1 ), start with the left side. first, substitute the values of ( cosh x ) and ( sinh x ) and use the formula ( (a b)^{2}=a^{2} b^{2} ).
( (cosh x)^{2}-(sinh x)^{2}=left\frac{1}{2}left(e^{x}+e^{-x}
ight)
ight^{2}-left\frac{1}{2}left(e^{x}-e^{-x}
ight)
ight^{2} )
( =\frac{1}{2}-\frac{1}{2} )
( =\frac{1}{4}-\frac{1}{4} )

Explanation:

Step1: Recall the property of the hyperbolic cosine function

The hyperbolic cosine function \(y = \cosh(x)=\frac{e^{x}+e^{-x}}{2}\) is an even function, i.e., \(\cosh(-x)=\cosh(x)\). Its graph is symmetric about the \(y -\)axis.

Step2: Analyze the given graphs

• Graph in option A: It is symmetric about the \(y -\)axis.
• Graph in option B: It is not symmetric about the \(y -\)axis (it is an odd - function - like graph, which is characteristic of the hyperbolic sine function \(y=\sinh(x)=\frac{e^{x}-e^{-x}}{2}\)).
• Graph in option C: The scale of the axes \([-6,6]\text{ by }[-1,6]\) is not relevant for the symmetry property. But the shape of the graph in option C is not symmetric about the \(y -\)axis in the correct proportion for the \(\cosh(x)\) function.

Since the function \(y = \cosh(x)\) is even and its graph is symmetric about the \(y -\)axis, the correct graph is the one that is symmetric about the \(y -\)axis.

Answer:

A.