Question

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

$f\left(x\right)=\sqrt{1+x}$,$g\left(x\right)=\sqrt{1-x}$

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$\left(x,y\right)$such that $y=f\left(x\right)$; that is, the graph of f is the graph of the $y=f\left(x\right)$$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$\left(x,y\right)$such that $y=f\left(x\right)$; that is, the graph of f is the graph of the $y=f\left(x\right)$$y=f\left(x\right)$

Step-2 – Analyze the given data

Given $f\left(x\right)=\sqrt{1+x}$, $g\left(x\right)=\sqrt{1-x}$

We first make a table of values.

x	$f\left(x\right)=\sqrt{1+x}$	$g\left(x\right)=\sqrt{1-x}$	$f\left(x\right)+g\left(x\right)$
-1	0	$\sqrt{2}$	$\sqrt{2}$
0	1	1	2
1	$\sqrt{2}$	0	$\sqrt{2}$

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$f\left(x\right)=\sqrt{1+x}$ is $D_f=\left\{x|x\ge -1\right\}$, $g\left(x\right)=\sqrt{1-x}$is $D_g=\left\{x|x\le 1\right\}$, the domain of$f\left(x\right)+g\left(x\right)$ is given by$D_{f+g}=D_f\cap D_g=\left\{x|-1\le x\le 1\right\}$