Chapter 1.8, Problem 26E

To determine

To find:Which function contributes most to the magnitude of the sum for both condition.

Answer to Problem 26E

For $0\le x\le 2$ , $g\left(x\right)$ contributes more to the sum than $f\left(x\right)$ , for $x>6$ , $f\left(x\right)$ contributes more to the sum than $g\left(x\right)$ .

Explanation of Solution

Given information:

The given conditions are $0\le x\le 2$ and $x>6$ .

The given functions are $f\left(x\right)=\frac{x}{2}$ and $g\left(x\right)=\sqrt{x}$ .

Calculation:

Letthe two functions $f$ and $g$ with overlapping domains.

For all $x$ common to both domains, the sum of $f$ and $g$ is given by, $\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)$ .

So,

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =\frac{x}{2}+\sqrt{x}\end{array}$

The graphs for the function $f\left(x\right)$ , $g\left(x\right)$ and $\left(f+g\right)\left(x\right)$ are plotted as shown in figure (1).

  

Figure (1)

From the graph shown in figure (1), it can be concluded that for $0\le x\le 2$ , $g\left(x\right)$ contributes more to the sum than $f\left(x\right)$ . For $x>6$ , $\left(f+g\right)\left(x\right)$ is positive, $f\left(x\right)$ is positive and $g\left(x\right)$ is positive. So, $f\left(x\right)$ contributes more to the sum than $g\left(x\right)$ .

Therefore, for $0\le x\le 2$ , $g\left(x\right)$ contributes more to the sum than $f\left(x\right)$ , for $x>6$ , $f\left(x\right)$ contributes more to the sum than $g\left(x\right)$ .