Chapter 4.2, Problem 49E

To determine

To show: The equation has only one solution in the given interval.

Answer to Problem 49E

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

Explanation of Solution

Given information:

The equation is:

  $x+\ln (x+1)=0$

  $0\le x\le 3$

Calculation:

The given equation is $x+\ln (x+1)=0$

Hence,

  $\begin{array}{c}0+\ln (0+1)=\ln 1\\ =0\end{array}$

Know that, $x+\ln (x+1)=0$ has at least one solution on the interval $(0,3)$ . To prove that $f(x)$ has exactly one, use a proof by contradiction:

Let $f(x)=x+\ln (x+1)$ , $f(x)$ is continuous and differentiable for all $x$ on its domain of $x>-1$ , it is continuous on the interval $(0,3)$ and differentiable on the interval $(0,3)$ . Assume that $f(x)$ has more than one solution for $f(x)=0$, on the interval $(0,3)$ where $x=0$ is one solution

  $x=a$ where $a\ne 0$ is another.

Therefore

  $\begin{array}{c}f(0)=f(a)\\ =0\end{array}$

The slopes between two these points is then

  $\begin{array}{c}\frac{f(a)-f(0)}{a-0}=\frac{0}{a}\\ =0\end{array}$

By the Mean Value Theorem, there must then be a $c$ in the interval $(0,a)$ such that $f^{\prime }(c)=0$

However

  $f\text{'}(x)=1+\frac{1}{x+1}>0$ in the interval $(0,3)$ that means $f^{\prime }(c)\ne 0$ for all $c$ in $(0,3)$

This contradicts that $f^{\prime }(c)=0$ for $(0,a)$ .

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