Chapter 2.4, Problem 46E

a.

To determine

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

Answer to Problem 46E

The answer is $f\left(x\right)=0$

Explanation of Solution

Given:

  $\ln x=3-2x$

Calculation:

The given equation is expressed as

  $\ln x+2x-3=0$ .

Consider the function $f\left(x\right)=\ln x+2x-3$

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

  $\underset{x\to -\infty }{\lim }f\left(x\right)=\underset{x\to -\infty }{\lim }\left(\ln x+2x-3\right)=-\infty$

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

  $\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\left(\ln x+2x-3\right)=\infty$

  $f\left(x\right)$ 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 46E

The answer is $\left(1.32,1.41\right)$ .

Explanation of Solution

Given:

  $length=0.01$

Calculation:

  $f\left(1.32\right)\approx 0.016>0$ and $f\left(1.41\right)\approx -0.014<0$

So, there is a root between 0.86 and 0.87 which is in the interval $\left(1.32,1.41\right)$ .