Chapter 2.4, Problem 48E

a.

To determine

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

Explanation of Solution

Given:

  $\sqrt{x-5}=\frac{1}{x+3}$

Calculation:

The equation is a continuous polynomial so we can use the Intermediate Value Theorem to prove the existence of a root. The Intermediate Value Theorem stats that if f is continuous on the close interval $\left[a,b\right]$ and let N be any number between $f\left(a\right)$ and $f\left(b\right)$ , where $f\left(a\right)$ does not equal $f\left(b\right)$ , then there exists a number c in $\left(a,b\right)$ such that $f\left(c\right)=N$ .

Using the theorem, let $a=5.01,b=5.02,and N=0$$f\left(a\right)=-0.0248andf\left(b\right)=0.0167$

N lies between $f\left(a\right)$ and $f\left(b\right)$

  $-0.0248<0<0.0167$

therefore, there exists a number c such that $f\left(c\right)=N$ .

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

b.

To determine

To find: the root corrects up to three decimal places by using the graph.

Explanation of Solution

Given:

  $\sqrt{x-5}=\frac{1}{x+3}$

Calculation:

The graph of $\sqrt{x-5}=\frac{1}{x+3}$ is given below:

So, from the graph theroot corrects up to three decimal places $5.016$ .