Question

Draw the graphs of f, g, and f + gon a common screen to illustrate graphical addition.

$f\left(x\right)=\sqrt[4]{1-x}$,$g\left(x\right)=\sqrt{1-\frac{x^2}{9}}$

Short answer

If f is a function with domain A, then the graph of f is the set of ordered pairs$\left\{\left(x,f\left(x\right)\right)|x\in A\right\}$ plotted in a coordinate plane. In other words, the graph of f is the set of all points ( x, y ) such that $y=f\left(x\right)$; that is, the graph of f is the graph of the $y=f\left(x\right)$

Step-by-step solution

Step-1 – Definition of graph of a function

If f is a function with domain A, then the graph of f is the set of ordered pairs$\left\{\left(x,f\left(x\right)\right)|x\in A\right\}$ plotted in a coordinate plane. In other words, the graph of f is the set of all points ( x, y ) such that $y=f\left(x\right)$; that is, the graph of f is the graph of the $y=f\left(x\right)$

Step-2 – Analyze the given data

Given$f\left(x\right)=\sqrt[4]{1-x}$,$g\left(x\right)=\sqrt{1-\frac{x^2}{9}}$

We first make a table of values.

x	$f\left(x\right)=\sqrt[4]{1-x}$	$g\left(x\right)=\sqrt{1-\frac{x^2}{9}}$	$f\left(x\right)+g\left(x\right)$
0	1	1	2
1	0	$\sqrt{\frac{8}{9}}$	$\sqrt{\frac{8}{9}}$

Step-3 – Graph

Plot the points given by the table and join them by a smooth curve to obtain the graph.

Since the domain of$f\left(x\right)=\sqrt[4]{1-x}$ is $D_f=\left\{x|x\le 1\right\}$, $g\left(x\right)=\sqrt{1-\frac{x^2}{9}}$is $D_g=\left\{x|x\ge -3\right\}$, the domain of$f\left(x\right)+g\left(x\right)$ is given by$D_{f+g}=D_f\cap D_g=\left\{x|-3\le x\le 1\right\}$