Chapter 3, Problem 83EP

(a).

To determine

The domain of the function.

Answer to Problem 83EP

The domain of the function is $\left\{x\in ℝ:-1<x<1\right\}$ .

Explanation of Solution

Given:

  $f\left(x\right)=\ln \left(1-x^2\right)$

Concept Used:

Domain of a function is the complete set of possible values of the independent variable of that function.

Calculation:

Graph of the function $f\left(x\right)=\ln \left(1-x^2\right)$ is below:

  

By analyzing the graph, the domain of the function is $\left\{x\in ℝ:-1<x<1\right\}$ .

Conclusion:

The domain of the function is $\left\{x\in ℝ:-1<x<1\right\}$ .

(b).

To determine

The value of $f\text{'}\left(x\right)$

Answer to Problem 83EP

The value of $f\text{'}\left(x\right)$ is $\frac{-2x}{\left(1-x^2\right)}$

Explanation of Solution

Given:

  $f\left(x\right)=\ln \left(1-x^2\right)$

Concept Used:

• The differentiation of $\frac{d}{dx}\left[\ln \left\{f(x)\right\}\right]=\frac{1}{f\left(x\right)}\cdot \frac{d}{dx}\left[f\left(x\right)\right]$ .

Calculation:

The derivative is as follow:

  $\begin{array}{c}f\text{'}\left(x\right)=\frac{d}{dx}\left[\ln \left(1-x^2\right)\right]\\ =\frac{1}{\left(1-x^2\right)}\cdot \frac{d}{dx}\left[1-x^2\right]\\ =\frac{\frac{d}{dx}\left[1\right]-\frac{d}{dx}\left[x^2\right]}{\left(1-x^2\right)}\\ =\frac{-2x}{\left(1-x^2\right)}\end{array}$

Conclusion:

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

(c).

To determine

To find the domain of $f\text{'}\left(x\right)$

Answer to Problem 83EP

The domain of $f\text{'}\left(x\right)$ is $\left(-\infty ,1\right)\cup \left(-1,1\right)\cup \left(1,\infty \right)$ .

Explanation of Solution

Given:

  $f\left(x\right)=\ln \left(1-x^2\right)$

Concept Used:

Domain of a function is the complete set of possible values of the independent variable of that function.

Calculation:

The derivative is $f\text{'}\left(x\right)=\frac{-2x}{\left(1-x^2\right)}$

The graph is below:

  

By analyzing the graph, the domain is:

  $x<-1\text{or}-1<x<1\text{or}x<1$ .

Conclusion:

The domain is $\left(-\infty ,1\right)\cup \left(-1,1\right)\cup \left(1,\infty \right)$

(d).

To determine

To show that $f\text{'}\text{'}\left(x\right)<0$ for all $x$ on the domain of $f$

Answer to Problem 83EP

  $f\text{'}\text{'}\left(x\right)<0$ for all $x$ on the domain of $f$

Explanation of Solution

Given:

  $f\left(x\right)=\ln \left(1-x^2\right)$

Concept Used:

• Domain of a function is the complete set of possible values of the independent variable of that function.
• $f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left[f\text{'}\left(x\right)\right]$

Calculation:

The first derivative from $\left(a\right)$ is $\frac{dy}{dx}=\frac{-2x}{\left(1-x^2\right)}$

Double derivative is:

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left[\frac{2x}{\left(x^2-1\right)}\right]\\ =2\cdot \frac{d}{dx}\left[\frac{x}{\left(x^2-1\right)}\right]\\ =2\cdot \frac{\left(x^2-1\right)\cdot \frac{d}{dx}\left[x\right]-x\cdot \frac{d}{dx}\left[x^2-1\right]}{{\left(x^2-1\right)}^2}\\ =2\cdot \frac{\left(x^2-1\right)-x\left[2x\right]}{{\left(x^2-1\right)}^2}\\ =\frac{2\left(-x^2-1\right)}{{\left(x^2-1\right)}^2}\\ =\frac{-2\left(x^2+1\right)}{{\left(x^2-1\right)}^2}\end{array}$

Conclusion:

  $\frac{d^{2}y}{d{x}^2}=\frac{-2\left(x^2+1\right)}{{\left(x^2-1\right)}^2}$ which is negative in $\left\{x\in ℝ:-1<x<1\right\}$ .