Chapter 4, Problem 83RE

a.

To determine

To find: the domain of f.

Answer to Problem 83RE

Domain of the function $\left|x\right|<1.$

Explanation of Solution

Given information: Given function is $f(x)=\ln (1-x^2).$

Calculation:

  $\ln z$ has a domain of $(0,\infty )\text{so}\ln (1-x^2)$ should have a domain that restricts $(1-x^2)$ to positive values.

  $\begin{array}{l}(1-x^2)>0.\\ 1>x^2.\\ 1>\left|x\right|.\end{array}$

b.

To determine

To find: f’(x).

Answer to Problem 83RE

  $f\text{'}(x)=\frac{2x}{x^2-1}.$

Explanation of Solution

Given information: Given function is $f(x)=\ln (1-x^2).$

Calculation:

  $\begin{array}{l}f(x)=\ln (1-x^2).\\ f\text{'}(x)=\frac{-2x}{1-x^2}.\\ f\text{'}(x)=\frac{-2x}{-(x^2-1)}=\frac{2x}{x^2-1}.\\ f\text{'}(x)=\frac{2x}{x^2-1}.\end{array}$

c.

To determine

To find: the domain of f’.

Answer to Problem 83RE

  $f\text{'}(x)=\frac{2x}{x^2-1}.$ is only undefined when x =-1 and x =1. The domain of

f(x)was restricted to (-1,1) though so the domain f’(x) must also be restricted to (-1,1).

Explanation of Solution

Given information: Given function is $f(x)=\ln (1-x^2).$

Calculation:

  $f\text{'}(x)=\frac{2x}{x^2-1}.$ is only undefined when x =-1 and x =1. The domain of

f(x)was restricted to (-1,1) though so the domain f’(x) must also be restricted to (-1,1).

d.

To determine

To prove: that f’’(x) <0 for all x in the domain of f.

Answer to Problem 83RE

Explanation of Solution

Given information: Given function is $f(x)=\ln (1-x^2).$

Calculation:

  $\begin{array}{l}f(x)=\ln (1-x^2).\\ f\text{'}(x)=\frac{-2x}{1-x^2}.\\ f\text{'}\text{'}(x)=\frac{(1-x^2)(-2)-(-2x)(-2x)}{{(1-x^2)}^2}.\\ f\text{'}\text{'}(x)=\frac{-2+2x^2-4x^2}{{(1-x^2)}^2}.\\ f\text{'}\text{'}(x)=\frac{-2-2x^2}{{(1-x^2)}^2}.\\ f\text{'}\text{'}(x)=\frac{-2(x^2+1)}{{(1-x^2)}^2}.\end{array}$

The domain of f is 0

In that domain (or for any $x\ne \pm 1$ ), the numerator of f’’ is always positive.

In the numerator, the $(x^2+1)$ part is always positive, but is multiplied by -2, which makes f’’(x)<0 in the domain of f.

Hence proved f’’(x) <0 for all x in the domain of f.