Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 20

Below is the graph of $f$


a.) Why is $f$ one-to-one?
b.) What are the domain and range of $f^{-1}$
c.) What is the value of $f^{-1}(2)$
d.) Estimate the use of $f^{-1}(0)$



a.) Based from the graph. The function is one-to-one since none of the $x$-values have the same $y$-values.
b.) If $f$ has a domain of $[-3,3]$ and range of $[-1,3]$. Thus, $f^{-1}$ has a domain of $[-1,3]$ and range of $[-3,3]$
c.) Let $x = f^{-1}(2)$, then $f(x) = f\left( f^{-1}(2) \right)$
$f(x) = 2$
At,
$x = 0$
Thus, $f^{-1}(2) = 0$

d.) Let $x = f^{-1}(0)$, then $f(x) = f\left( f^{-1}(0)\right)$
$f(x) = 0$
At,
$x \approx - 1.8$
Thus, $f^{-1}(0) \approx -1.8$