# Single Variable Calculus, Chapter 7, 7.1, Section 7.1, Problem 36

Suppose that $\displaystyle f(x) = \frac{1}{x-1}$, $x > 1$, $a = 2$

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 $\displaystyle f(x) = \frac{1}{x-1}$, then

$\displaystyle f'(x) = \frac{-1}{(x-1)^2} < 0$ for its domain $(1, \infty)$

$f$ is always decreasing, thus, no values of $x$ will give the same values of $y$. Therefore, $f$ is a one-to-one.

b.) Based from the theorem,

\begin{aligned} \left( f^{-1} \right)' (a) &= \frac{1}{f'\left( f^{-1}(a) \right)}\\ \\ \text{it reflect } x = f^{-1}(2), \text{ then}\\ \\ f(x) &= f\left( f^{-1} (2) \right)\\ \\ \frac{1}{x-1} &= 2\\ \\ 1 &= 2x - 2\\ \\ 2x &= 3 \\ \\ x &= \frac{3}{2} \end{aligned}

So, $\displaystyle f^{-1}(2) = \frac{3}{2}$

Thus,

\begin{aligned} \left(f^{-1} \right)'(2) &= \frac{1}{f'\left( \frac{3}{2}\right)} = \frac{1}{\frac{-1}{(x-1)^2}} = -(x-1)^2\\ \\ &= - \left( \frac{3}{2} - 1 \right)^2 = -\frac{1}{4} \end{aligned}

c.) If $\displaystyle f(x) = \frac{1}{x-1}$, then

\begin{aligned} f^{-1} (x) \quad \Longrightarrow \quad x &= \frac{1}{y-1}\\ \\ y-1 &= \frac{1}{x}\\ \\ y &= \frac{1}{x} +1 \end{aligned}

Thus,

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

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

$\displaystyle \left( f^{-1} \right)'(x) = \frac{-1}{x^2}$

when $x = 2$

$\displaystyle \left( f^{-1} \right)'(2) = \frac{-1}{2^2} = \frac{-1}{4}$

We can say that our answers agree with part(b)

e.)

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