Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises - Page 345: 89

False;

An even function is such that $f(x) = f(-x)$, $\forall x∈\mathcal D$. So, an even function is never one-to-one for every point in the range* has at least two points as its preimages. Thus $f$ is not one-to-one and hence $f^{-1}$ does not exist. For example, if $f(x)=x^2+4$ , which is an even function, $f^{-1}$ does not exist. *except for f(0)