Chapter 2.4, Problem 45E

a.

To determine

To prove:that the given equation has only one root.

Answer to Problem 45E

  $f\left(x\right)$ takes the value $-\infty \text{ and }+\infty$ . The value of for $f\left(x\right)=0$ .

Explanation of Solution

Given:

  $\cos x=x^3$

Calculation:

The given equation is expressed as

  $0=x^3-\cos x$ .

Consider the function $f\left(x\right)=x^3-\cos x$

Now,taking $\underset{x\to -\infty }{\lim }$ on both the sides,

  $\begin{array}{l}\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(x^3-\cos x\right)\\ \underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }{x}^3-\underset{x\to -\infty }{\lim }\cos x=-\infty \end{array}$

Again, taking $\underset{x\to \infty }{\lim }$ on both the sides,

  $\begin{array}{l}\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(x^3-\cos x\right)\\ \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }{x}^3-\underset{x\to \infty }{\lim }\cos x=\infty \end{array}$

  $f\left(x\right)$ is continuous function for all x .

According to the intermediate value theorem, $f\left(x\right)$ will take all values between $-\infty \text{ and }+\infty$ .

Therefore, there is at least one value for x for which $f\left(x\right)$ is zero.

Hence, there is one root for $f\left(x\right)=0$ .

b.

To determine

To find: the interval so that the length 0.01 that contains a root.

Answer to Problem 45E

The answer is $\left(0.86,0.87\right)$

Explanation of Solution

Given:

  $length=0.01$

Calculation:

  $f\left(0.86\right)\approx 0.016>0$ and $f\left(0.87\right)\approx -0.014<0$

So, there is a root between 0.86 and 0.87 which is in the interval $\left(0.86,0.87\right)$ .