Chapter 1.8, Problem 25E

To determine

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

Answer to Problem 25E

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

Explanation of Solution

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

The given functions are $f\left(x\right)=3x$ and $g\left(x\right)=-\frac{x^3}{10}$ .

Calculation:

Let two functions $f$ and $g$ are in 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,

  $\begin{array}{c}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\\ =3x-\frac{x^3}{10}\\ =-\frac{x^3}{10}+3x\end{array}$

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 $0\le x\le 2$ , $f\left(x\right)$ contributes more to the sum than $g\left(x\right)$ .For $x>6$ , $\left(f+g\right)\left(x\right)$ is negative, $f\left(x\right)$ is positive and $g\left(x\right)$ is negative. So, $g\left(x\right)$ contributes more to the sum than $f\left(x\right)$ .

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