Chapter 5.2, Problem 49E

To determine

To show : That the equation has exactly one solution in the interval.

Answer to Problem 49E

The possible explanation is given below.

Explanation of Solution

Given information :

The given equation is $x+\ln \left(x+1\right)=0,0\le x\le 3$

Calculation :

Let $f\left(x\right)=x+\ln \left(x+1\right)$ .

Substitute $x=0$ and solve for $f\left(x\right)$ :

  $\begin{array}{l}f\left(0\right)=0+\ln \left(0+1\right)\\ f\left(0\right)=\ln \left(1\right)\\ f\left(0\right)=0\end{array}$

Since, it is known that $f\left(x\right)$ has at least one solution on the interval $\left[0,3\right]$ .

To prove that $f\left(x\right)$ has exactly one solution use proof by contradiction.

Since, $f\left(x\right)$ is continuous on the interval $\left[0,3\right]$ and differentiable on the interval $\left(0,3\right)$ .

Suppose, that $f\left(x\right)$ has one more solution for $f\left(x\right)=0$ on the interval $\left[0,3\right]$ , where $x=0$ is one solution and $x=t,$ where $t=0$ is another.

Therefore, $f\left(0\right)=f\left(a\right)=0$ .

The slope between the two point is

  $\begin{array}{l}\frac{f\left(a\right)-f\left(0\right)}{a-0}=\frac{0}{a-0}\\ \frac{f\left(a\right)-f\left(0\right)}{a-0}=0\end{array}$

Thus, by Mean Value Theorem, there must be a $c$ in the interval $\left(0,a\right)$ such that $f\text{'}\left(c\right)=0$ .

However, $f\text{'}\left(x\right)=1+\frac{1}{x+1}>0$ in the interval in the interval $\left[0,3\right]$ , which means $f\text{'}\left(c\right)\ne 0$ for all $c$ in $\left(0,3\right)$ . This contradicts the fact that $f\text{'}\left(c\right)=0$ for $\left(0,a\right)$ .

Therefore ,

  $f\left(x\right)$ cannot have a second solution for $f\left(x\right)=0$ so $x+\ln \left(x+1\right)=0$ has exactly one solution.