Chapter 3, Problem 15TP

Solution

The x -values for which $f\left(x\right)=g\left(x\right)$ .

  $x=\frac{19}{9}$

Given:

The functions $f\left(x\right)=\sqrt{x-2}+3\text{ and }g\left(x\right)=\sqrt{x+13}$ .

Concept Used:

• Power of a power property: ${\left(x^a\right)}^b=x^{a\cdot b}$
• Product of power property: $a^m\cdot a^n=a^{m+n}$
• Quotient of power property: $\frac{a^m}{a^n}=a^{m-n}$
• Product property of radicals: $\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$
• Quotient property of radicals: $\frac{\sqrt[n]{a}}{\sqrt[n]{b}}=\sqrt[n]{\frac{a}{b}}$
• Sum of like radicals’ property: $p\sqrt[n]{a}+q\sqrt[n]{a}=\left(p+q\right)\sqrt[n]{a}$
• The exponent form of the nth root of a number a is given by $\sqrt[n]{a}=a^{\frac{1}{n}}$

Calculation:

In order to find the x -values for which $f\left(x\right)=g\left(x\right)$ , equate the given functions and simplify further as shown below

  $\begin{array}{c}f\left(x\right)=g\left(x\right)\\ \sqrt{x-2}+3=\sqrt{x+13}\\ {\left(\sqrt{x-2}+3\right)}^2={\left(\sqrt{x+13}\right)}^2\\ \left(x+2\right)+6\sqrt{x-2}+9=x+13\\ x+6\sqrt{x-2}+11=x+13\\ 6\sqrt{x-2}=2\\ \sqrt{x-2}=\frac{2}{6}\\ \sqrt{x-2}=\frac{1}{3}\\ {\left(\sqrt{x-2}\right)}^2={\left(\frac{1}{3}\right)}^2\\ x-2=\frac{1}{9}\\ x=\frac{1}{9}+2\\ x=\frac{1+18}{9}\\ x=\frac{19}{9}\end{array}$

Thus, the value of x for which $f\left(x\right)=g\left(x\right)$ is $x=\frac{19}{9}$ .