## Calculus 8th Edition

a). $f(x)=\sqrt{ x-2}$ is one-to-one because it never takes on the same value twice, i.e. $f(x_{1})\ne(x_{2})$ whenever $x_{1}\ne x_{2}$. Furthermore, by the horizontal line test it is also one-to-one. b).$(f^{-1})'(2)=4$. c). $f^{-1}(x)=x^{2}+2$. This domain is $(-\infty,\infty)$, the range is also $(-\infty,\infty)$. d). $(f^{-1})'(2)=4$ e). See the attached graph. The grey line in the attached graph represents $f$ and the purple line represents $f^{-1}$
a). In the attached graph, the grey line represents $f$, and the red and green lines represent $f(x)=1$ and $f(x)=-1$ respectively. From the horizontal line test, we can clearly see that $f$ is one-to-one because the horizontal line intersects its graph only once. b). We are given $f(x)=\sqrt{ x-2}$, so $f'(x)=\frac{1}{2}(x-2)^{-\frac{1}{2}}$. To find the inverse of $f$, let $y=\sqrt{ x-2}$ then solving in terms of $x$, we get $x=y^{2}+2$. Now swap the x and y terms, so $y=f^{-1}(x)=x^{2}+2$. Substitute $x=a=2$ into $f^{-1}$ to get $f^{-1}(2)=6$. Now, by theorem 7, $(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(6)}=\frac{1}{\frac{1}{2}(6-2)^{-\frac{1}{2}}}=\frac{1}{\frac{1}{4}}=4$. c). From the above, $f^{-1}(x)=x^{2}+2$. The domain and range of this function is all real numbers, we can see this from the graph of $f^{-1}(x)$, which is a parabola that has a y-intercept at x=0,y=2. d). From c, $(f^{-1})'(x)=2x$. So, $(f^{-1})'(2)=4$. e). The grey line in the attached graph represents $f$ and the purple line represents $f^{-1}$