What kind of symmetry is displayed by the graph of catenary y=e^x+e^-x/2
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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:
[tex]f(x) = \frac{e^x + e^{-x}}{2}[/tex]
We make:
[tex]f(-x) = \frac{e^{-x} + e^{-(-x)}}{2}[/tex]
Rewriting the function we have
[tex]f(-x) = \frac{e^{-x} + e^{x}}{2} = \frac{e^{x} + e^{-x}}{2}\\\\f(-x) = f(x)[/tex]
Finally, the function has symmetry with respect to the y axis.