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\SN{6.1}
\pn13.
$y=x^2+1$, $x^2=y-1$, $x=\pm\sqrt{y-1}$,  so
$f^{-1}(x)=\sqrt{x-1}$ and we want the plus sign because
the domain of $f$ is $x\geq 0$.

\pn15. $y=x^3-1$, $x^3=y+1$, $x=\root{3}\of{y+1}$, so
$f^{-1}(t)=\root 3\of{t+1}$.

\pn23. $y=\frac 1{x^2}$, $x^2=\frac 1y$, $x=\frac {\pm1}{\sqrt y}$, so
$f^{-1}(x)=\frac 1{\sqrt x}=x^{-1/2}$ since the domain of $f$ is $x>0$.
The domain of $f^{-1}$ is all $x>0$; the range of $f^{-1}$ is all
$x>0$.
These follow from the corresponding results for $f$.
For $f$, we were told that the domain is $x>0$.
This gives the range of $f^{-1}$.
By definition the range consists of all numbers $y$ so
that we can solve the equation $\frac 1{x^2}=y$ with $x>0$.
This in turn is all $y$, $y>0$, which gives the range of $f$ and
the domain of $f^{-1}$.

\def\dfinv{df\hbox to 0pt {$\scriptstyle^{-1}$\hss}}

\pn31. $f(x)=x^3-3x^2-1$, $x\geq 2$; 
find $\frac{\dfinv}{dx}$\hskip10pt at $x=-1$.
Formula: $\frac{\dfinv}{dx}\hskip8pt(a)=
\frac{1}{f^\prime\br1{(f^{-1}(a)}}$.
We want $\frac{\dfinv}{dx}\hskip8pt(-1)$ so what is 
$f^{-1}(-1)$?
Equivalently, solve $f(a)=-1$ and we are told that the answer is $3$.
Since $f^\prime(x)=3x^2-6x$,
$\frac{\dfinv}{dx}\hskip8pt(-1)=\frac 1{3(3^2)-6\cdot3}=
\frac 19$.

\pn33. This problem is another application of 
the formula in the last problem.
Since $(2,4)$ is on the graph of $f$, $(4,2)$ is on the graph of $f^{-1}$,
or $f^{-1}(4)=2$.
The slope of $f$ at $2$ is $\frac 13$ and 
you may proceed either of two ways.
Algebraic: $f^\prime(2)=\frac 13$;
$\frac{\dfinv}{dx}\hskip8pt(4)=\frac 1{f^\prime(2)}$ since
$f^{-1}(4)=2$.  
Hence $\frac{\dfinv}{dx}\hskip8pt(4)=3$.
Geometric: 1 equals the slope of a line times
the slope of that line reflected in $y=x$.
Hence $\Bigl(\frac{\dfinv}{dx}\hskip8pt(4)\Bigr)\cdot f^\prime(2)=1$.

\pn43. $f(x)=(1-x)^3$; $f^\prime(x)=-3(1-x)^2$.
Hence $f^\prime(x)\leq 0$ and is equal to $0$ only if $x=1$.
Hence $f$ is decreasing and therefore one to one.
$\frac{\dfinv}{dx}\hskip8pt(x)=
\frac 1{f^\prime\bigl(f^{-1}(x)\bigr)}=
\frac 1{-3\bigl(1-f^{-1}(x)\bigr)^2}$.
To continue, we need a formula for $f^{-1}$:
$y=(1-x)^3$, $\root 3\of y=1-x$, $x=1-\root3\of y$,
so $f^{-1}(x)=1-\root3\of x$.
Hence 
$\frac{\dfinv}{dx}\hskip8pt(x)=
\frac 1{-3\bigl(1-(1-\root3\of x\>)\bigr)^2}=
\frac 1{-3(\root3\of x\>)^2}=\frac 1{-3x^{2/3}}=-\frac 13 x^{-2/3}$.


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