Chapter 3, Problem 1RCC

a)

To determine

Define one to one function.

How can conclude the function as one-to-one by looking at the graph of the function.

Explanation of Solution

One to one function:

When a function does not takes the same value twice, then the function is called as one to one function.

$f\left(x_1\right)\ne f\left(x_2\right)\text{whenever}{x}_1\ne x_2$

A function is said to be one to one function when no horizontal line intersects its graph more than once. So, the function is said to be as one to one by looking the graph.

b)

To determine

Define the inverse function $f^{-1}$ if f is a one-to-one function.

Provide the explanation to obtain the graph of the function $f^{-1}$ from the graph of the function f.

Explanation of Solution

Consider the function f has a domain as a and range as b. Then, the domain and range of the function $f^{-1}$ are b and a.

The graph of inverse functions are reflections over the line $y=x$. This means that each x-value must be matched to one and only one y-value.

Functions which meet the above criteria are called one to one function.

$\begin{array}{l}{f}^{-1}\left(y\right)=x\\ f\left(x\right)=y\end{array}$

c)

To determine

Find the formula for ${\left(f^{-1}\right)}^{\prime }\left(a\right)$ if the value of $f^{\prime }\left(f^{-1}\left(a\right)\right)\ne 0$.

Answer to Problem 1RCC

The formula for ${\left(f^{-1}\right)}^{\prime }\left(a\right)$ is ${\left(f^{-1}\right)}^{\prime }\left(a\right)=\frac{1}{f^{\prime }\left[f^{-1}\left(a\right)\right]}$.

Explanation of Solution

Given information:

The function f is a one to one function.

Calculation:

Consider the function f is one to one differentiable function with the inverse function $f^{\prime }$.

Show the relation as follows:

$f^{\prime }\left(f^{-1}\left(a\right)\right)\ne 0$

The inverse function is differentiable at a.

${\left(f^{-1}\right)}^{\prime }\left(a\right)=\frac{1}{f^{\prime }\left[f^{-1}\left(a\right)\right]}$