Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 34

Suppose that $f(x) = \sqrt{x-2}$, $a = 2$
a.) Show that $f$ is a one-to-one.
b.) Use the theorem in inverse function to find for $(f^{-1})' (a)$
c.) Calculate $f^{-1}(x)$ and state the domain and range of $f^{-1}$
d.) Calculate $(f^{-1}) (a)$ from the formula in part(c) and check that it agrees with the results in part(b).
e.) On the same plane, sketch the graphs of $f$ and $f^{-1}$

a.) If $f(x) = \sqrt{x-2}$. Then,
$\displaystyle f'(x) = \frac{1}{2\sqrt{x - 2}} > 0$ for its doamion $[2, \infty)$
$f$ is always increasing, therefore, no values of $x$ will give the same values of $y$. Thus, $f$ is a one-to-one.

b.) Based from the theorem,

$
\begin{equation}
\begin{aligned}
\left( f^{-1} \right)' (x) &= \frac{1}{f' \left( f^{-1}(a) \right)}\\
\\
\text{we know that } f'(x) &= \frac{1}{2\sqrt{x-2}}\\
\\
\text{if we let } x = f^{-1}(2), \text{ then}\\
\\
f(x) &= f\left( f^{-1}(2) \right)\\
\\
\sqrt{x-2} &= 2\\
\\
x - 2 &= \pm 4\\
\\
x &= \pm 4 + 2
\end{aligned}
\end{equation}
$

We got $x = 6$ and $x = -2$, however $x = -2$ is not defined in the domain of $x$. Therefore, $f^{-1}(2) = 6$
Thus,

$
\begin{equation}
\begin{aligned}
\left( f^{-1} \right)'(2) &= \frac{1}{f'\left( f^{-1}(2) \right)} = \frac{1}{f'(6)} = \frac{1}{\frac{1}{2\sqrt{6-2}}}\\
\\
&= 2\sqrt{4} = 4
\end{aligned}
\end{equation}
$

c.) If $f(x) = \sqrt{x-2}$, then

$
\begin{equation}
\begin{aligned}
f^{-1}(x) \quad \Longrightarrow \quad x &= \sqrt{y - 2}\\
\\
x^2 &= y - 2\\
\\
y &= 2 + x^2
\end{aligned}
\end{equation}
$

Thus,
$f^{-1}(x) = 2 + x^2$, we know that the domain of $f$ is $[2, \infty)$ and its range is $[0, \infty)$. Thus, the domain of $f^{-1}(x)$ is $[0,\infty)$ and its range is $[2, \infty)$.

d.) If $f^{-1}(x) = 2 + x^2$, then

$
\begin{equation}
\begin{aligned}
\left( f^{-1} \right)' (x) &= 2x \\
\\
\text{when } x = 2, \text{ then}\\
\\
\left( f^{-1} \right)'(2) &= 2(2)\\
\\
\left( f^{-1} \right)' (2) &= 4
\end{aligned}
\end{equation}
$

We can say that the answer agree with part(b)

e.)

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