function's symmetry

Question

I have the function $f(x)=e^x+e^{2-x},x\in\mathbb{R}$ and I want to prove that $f$'s graph has symmetry with respect to $x=1$. Any ideas? Can I prove that the function $f(x-1)$ is odd?

Answer 1

You have to prove that $f(1-x)=f(1+x)$ (which is the same as $x \mapsto f(1-x)$ is even)

Here you have: $$f(1-x)=e^{1-x}+e^{2-(1-x)}=e^{1-x}+e^{1+x}$$ $$f(1+x)= e^{1+x}+e^{2-(1+x)}=e^{1+x}+e^{1-x}$$

Answer 2

Let consider the horizontal translation $X+1=x$

• $f(x)=e^x+e^{2-x}\to f(X)=e^{1+X}+e^{1-X}$

and

• $f(X)=f(-X)$