## Calculus 8th Edition

a). $f(x)=\frac{1}{x-1}$ is one-to-one for $x\gt 1$ because it never takes on the same value twice in this domain, 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 in this domain. b).$(f^{-1})'(2)=-\frac{1}{4}$. c). $f^{-1}(x)=\frac{1}{x} + 1$. This domain is $(-\infty,0)$U$(0,\infty)$, the range is also $(-\infty,1)$U$(1,\infty)$. d). $(f^{-1})'(2)=-\frac{1}{4}$ e). See the attached graph. The red line represents $f^{-1}$, the blue line represents $f$ and the purple line represents the horizontal line $x=1$.
a). The blue line in the attached graph represents $f(x)=\frac{1}{x-1}$, the red and purple lines represent the horizontal lines f(x)=1 and f(x)=-1 respectively. From the horizontal line test, we can clearly see that the function is not one-to-one over the whole domain of x, however for $x \ge 1$ no horizontal line intersects its graph more than once. b). We are given $f(x)=\frac{1}{x-1}$, so $f'(x)=-\frac{1}{{(x-1)}^{2}}$. To find the inverse of $f$, let $y=\frac{1}{x-1}$ then solving in terms of $x$, we get $x=\frac{1}{y} + 1$. Now swap the x and y terms, so $y=f^{-1}(x)=\frac{1}{x} + 1$. Substitute $x=a=2$ into $f^{-1}$ to get $f^{-1}(2)=\frac{3}{2}$. Now, by theorem 7, $(f^{-1})'(2)=\frac{1}{f'(f^{-1}(2))}=\frac{1}{f'(\frac{3}{2})}=-\frac{1}{\frac{1}{\frac{1}{2^{2}}}}=-\frac{1}{4}$. c). From the above, $f^{-1}(x)=\frac{1}{x} + 1$. So at $x=0$, the function is undefined. Furthermore, the function tends to $-\infty$ as $x$ tends to zero from below and it tends to $+\infty$ as $x$ tends to zero from above. d). From c, $(f^{-1})'(x)=-\frac{1}{x^2}$. So, $(f^{-1})'(2)=-\frac{1}{4}$. e). See the attached graph. The red line represents $f^{-1}$, the blue line represents $f$ and the purple line represents the horizontal line $x=1$.