Chapter 5, Problem 9RE

a.

To determine

To find the intervals on which the function is increasing by using analytical method.

Answer to Problem 9RE

There are no values of $x$ for which $y={\cos }^{-1}x$ is increasing.

Explanation of Solution

Given:

The function is $y={\cos }^{-1}x$ .

Calculation:

The function is increasing when $f^{\prime }\left(x\right)>0$ .

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

Since $\sqrt{1-x^2}>0$

Therefore, $\frac{1}{\sqrt{1-x^2}}$ is a positive term

So, $-\frac{1}{\sqrt{1-x^2}}<0$

It means $f^{\prime }\left(x\right)<0$ , $y={\cos }^{-1}x$ is always decreasing function for all values of $x$ in its domain $\left[-1,1\right]$ .

Below is graph of function $y={\cos }^{-1}x$

  

From above graph it is clear that function $y={\cos }^{-1}x$ is always decreasing when $\left[-1,1\right]$

Hence , there are no values of $x$ for which $y={\cos }^{-1}x$ is increasing

b.

To determine

To find the intervals on which the function is decreasing by using analytical method.

Answer to Problem 9RE

  $y={\cos }^{-1}x$ is always decreasing function for all values of $x$ in its domain $\left[-1,1\right]$ .

Explanation of Solution

Given:

The function is $y={\cos }^{-1}x$ .

Calculation:

The function is decreasing when $f^{\prime }\left(x\right)<0$ .

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

Since $\sqrt{1-x^2}>0$

Therefore, $\frac{1}{\sqrt{1-x^2}}$ is a positive term

So, $-\frac{1}{\sqrt{1-x^2}}<0$

It means $f^{\prime }\left(x\right)<0$ , $y={\cos }^{-1}x$ is always decreasing function for all values of $x$ in its domain $\left[-1,1\right]$ .

Below is graph of function $y={\cos }^{-1}x$

  

From above graph it is clear that function $y={\cos }^{-1}x$ is always decreasing when $\left[-1,1\right]$

c.

To determine

To find the intervals on which the function is concave up by using analytical method.

Answer to Problem 9RE

The graph of function is concave up when $x\in \left(-1,0\right)$

Explanation of Solution

Given:

The function is $y={\cos }^{-1}x$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y={\cos }^{-1}x$

First derivative : $y^{\prime }=-\frac{1}{\sqrt{1-x^2}}$

Second derivative : $y^{″}=\frac{-x}{{\left(1-x\right)}^{3/2}}$

Now, put $f^{″}\left(x\right)=0$ to find critical points

  $\begin{array}{l}\frac{-x}{{\left(1-x\right)}^{3/2}}=0\\ \therefore x=0\text{ is one of critical point}\text{.}\end{array}$

Domain of $y={\cos }^{-1}x$ is $\left[-1,1\right]$

Therefore, there are two intervals that is $\left(-1,0\right)$ and $\left(0,1\right)$

check the value of $f^{″}\left(x\right)$ in interval $\left(-1,0\right)$ and $\left(0,1\right)$

Now for $x\in \left(-1,0\right)$ test for $x=-0.5$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-x}{{\left(1-x\right)}^{3/2}}\\ f^{″}\left(-0.5\right)=\frac{-\left(-0.5\right)}{{\left(1-\left(-0.5\right)\right)}^{3/2}}=0.272>0\end{array}$

Hence $f^{″}\left(x\right)>0$ when $x\in \left(-1,0\right)$ ,it means that the graph of function is concave up when $x\in \left(-1,0\right)$

Below is graph of function $y={\cos }^{-1}x$

  

From above graph it is clear that function $y={\cos }^{-1}x$ is concave up when $x\in \left(-1,0\right)$

d.

To determine

To find the intervals on which the function is concave down by using analytical method.

Answer to Problem 9RE

The graph of function is concave down when $x\in \left(0,1\right)$

Explanation of Solution

Given:

The function is $y={\cos }^{-1}x$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y={\cos }^{-1}x$

First derivative : $y^{\prime }=-\frac{1}{\sqrt{1-x^2}}$

Second derivative : $y^{″}=\frac{-x}{{\left(1-x\right)}^{3/2}}$

Now, put $f^{″}\left(x\right)=0$ to find critical points

  $\begin{array}{l}\frac{-x}{{\left(1-x\right)}^{3/2}}=0\\ \therefore x=0\text{ is one of critical point}\text{.}\end{array}$

Domain of $y={\cos }^{-1}x$ is $\left[-1,1\right]$

Therefore, there are two intervals that is $\left(-1,0\right)$ and $\left(0,1\right)$

On checking value of $f^{″}\left(x\right)$ in interval $\left(-1,0\right)$ and $\left(0,1\right)$

Now for $x\in \left(0,1\right)$ test for $x=0.5$

  $\begin{array}{l}{f}^{″}\left(x\right)=\frac{-x}{{\left(1-x\right)}^{3/2}}\\ f^{″}\left(0.5\right)=\frac{-\left(0.5\right)}{{\left(1-\left(0.5\right)\right)}^{3/2}}=-1.41<0\end{array}$

Hence $f^{″}\left(x\right)<0$ when $x\in \left(0,1\right)$ ,it means that the graph of function is concave down when $x\in \left(0,1\right)$

Below is graph of function $y={\cos }^{-1}x$

  

From above graph it is clear that function $y={\cos }^{-1}x$ is concave down when $x\in \left(0,1\right)$

e.

To determine

To find any local extreme values.

Answer to Problem 9RE

  $y={\cos }^{-1}x$ has local maximum value at point $\left(-1,\pi \right)$ and local minimum value at point $\left(1,0\right)$

Explanation of Solution

Given:

The function is $y={\cos }^{-1}x$ .

Calculation:

Graph of $y={\cos }^{-1}x$ is given below

  

From graph it is clear that $y={\cos }^{-1}x$ has local maximum value at point $\left(-1,\pi \right)$ and local minimum value at point $\left(1,0\right)$ .

f.

To determine

To find inflections points.

Answer to Problem 9RE

The inflection point of $y={\cos }^{-1}x$ is $\left(0,\frac{\pi }{2}\right)$ .

Explanation of Solution

Given:

The function is $y={\cos }^{-1}x$ .

Calculation:

Inflection point of any function is a point where the graph of function has a tangent line and where the concavity changes.

Graph of $y={\cos }^{-1}x$ is given below

  

From graph it is clear that concavity of $y={\cos }^{-1}x$ changes at $\left(0,\frac{\pi }{2}\right)$ therefore

The inflection point of $y={\cos }^{-1}x$ is $\left(0,\frac{\pi }{2}\right)$ .