Question

What kind of symmetry is displayed by the graph of catenary y=e^x+e^-x/2

Answer 1

Answer:

Option b

Step-by-step explanation:

To solve this problem we must test if the function is even.

If f(-x) = f(x) then the function is even and is symmetric with respect to the y-axis.

If f(-x) = -f(x) then the function is odd and has symmetry with respect to the origin.

We have the function:

$f(x) = \frac{e^x + e^{-x}}{2}$

We make:

$f(-x) = \frac{e^{-x} + e^{-(-x)}}{2}$

Rewriting the function we have

$f(-x) = \frac{e^{-x} + e^{x}}{2} = \frac{e^{x} + e^{-x}}{2}\\\\f(-x) = f(x)$

Finally, the function has symmetry with respect to the y axis.