Calculus 8th Edition

a). $f(x)=x^{3}$ 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).$(f^{-1})'(8)=\frac{1}{12}$. c). $f^{-1}(x)=x^{\frac{1}{3}}$. This domain is $(-\infty,\infty)$, the range is also $(-\infty,\infty)$. d). $(f^{-1})'(8)=\frac{1}{12}$ e). See the attached graph. The grey line in the attached graph represents $f$ and the green line represents $f^{-1}$
a). The green line in the attached graph represents $x^{3}$, the red and grey 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 one-to-one because no horizontal line intersects its graph more than once. b). We are given $f(x)=x^{3}$, so $f'(x)=3x^{2}$. To find the inverse of $f$, let $y$=$x^{3}$ then solving in terms of $x$, we get $x=y^{\frac{1}{3}}$. Now swap the x and y terms, so $y=f^{-1}(x)=x^{\frac{1}{3}}$. Substitute $x=a=8$ into $f^{-1}$ to get $f^{-1}(8)=2$. Now, by theorem 7, $(f^{-1})'(8)=\frac{1}{f'(f^{-1}(8))}=\frac{1}{f'(2)}=\frac{1}{3\times2^{2}}=\frac{1}{12}$. c). From the above, $f^{-1}(x)=x^{\frac{1}{3}}$. The domain and range of this function is all real numbers, the cube root of any number is either positive, negative or zero. d). From c, $(f^{-1})'(x)=\frac{1}{3}x^{-\frac{2}{3}}$. So, $(f^{-1})'(8)=\frac{1}{3}8^{-\frac{2}{3}}=\frac{1}{3}\frac{1}{4}=\frac{1}{12}$. e). The grey line in the attached graph represents $f$ and the green line represents $f^{-1}$