## Calculus 8th Edition

a). $f(x)=9-x^{2}$ is not a one-to-one because it can takes on the same value twice, i.e. $f(x_{1})=f(x_{2})$ at some values of $x_{1}$ and $x_{2}$ e.g. when $x_{1}=2$ and $x_{2}=-2$, then $f(x_{1})=f(x_{2})$. However, on the domain $[0,3]$ as given in the question, it is one-to-one. Furthermore, the horizontal line test of this function in the domain $0\leq x \leq 3$ shows it is one-to-one. b).$(f^{-1})'(8)=-\frac{1}{2}$. c). $f^{-1}(x)=\sqrt {9-x}$. This domain is $(-\infty,9]$, the range is $[0,\infty)$. d). $(f^{-1})'(8)=-\frac{1}{2}$ e). See the attached graph. The grey line in the attached graph represents $f^{-1}$ and the green line represents $f$
a). In the attached graph, the green line represents $f$, and the blue and orange lines represent $f(x)=1$ and $f(x)=-1$ respectively. From the horizontal line test, we can clearly see that $f$ is not a one-to-one function because the horizontal line intersects its graph more than once. However in the domain $[0,3]$ it is one-to-one. b). The function is one-to-one in the domain $[0,3]$ of x. So, We are given $f(x)=9-x^{2}$, then $f'(x)=-2x$. To find the inverse of $f$, let $y=9-x^{2}$ then solving in terms of $x$, we get $x=\sqrt {9-y}$. Now swap the x and y terms, so $y=f^{-1}(x)=\sqrt {9-x}$. Substitute $x=a=8$ into $f^{-1}$ to get $f^{-1}(8)=1$. Now, by theorem 7, $(f^{-1})'(8)=\frac{1}{f'(f^{-1}(8))}=\frac{1}{f'(1)}=\frac{1}{-2\times1}=-\frac{1}{2}$. c). We are given $f(x)=9-x^{2}$. To find the inverse of $f$, let $y=9-x^{2}$ then solving in terms of $x$, we get $x=\sqrt {9-y}$. Now swap the x and y terms, so $y=f^{-1}(x)=\sqrt {9-x}$. From the graph of this inverse function, we can see that the function is undefined for values of x greater than 9. d). From c, $(f^{-1})'(x)=-\frac{1}{2}(9-x)^{-\frac{1}{2}}$. So, $(f^{-1})'(8)=-\frac{1}{2}$. e). The grey line in the attached graph represents $f^{-1}$ and the green line represents $f$