Chapter 2, Problem 61RE

To determine

To sketch: the graph of the function. Also state the domain, intercepts and range of the graph.

Answer to Problem 61RE

The intercepts are $\left(1,0\right),\left(0,1\right)$ .

The domain is $\left\{x|x\le 1\right\}$ or the interval $\left[-\infty ,1\right]$ .

The intercepts are $\left(1,0\right),\left(0,1\right)$ , domain is $\left\{x|x\le 1\right\}$ or the interval $\left[-\infty ,1\right]$ , and range is $\left\{y|y\ge 0\right\}$ or the interval $\left[0,\infty \right)$ .

Explanation of Solution

Given information:

Given function

  $f\left(x\right)=\sqrt{1-x}$

Calculation:

Graph the basic function $f\left(x\right)=\sqrt{x}$ .

  

Plot the graph $f\left(x\right)=\sqrt{-x}$ by replacing $x$ with $-x$ .

  

Shift the graph of $f\left(x\right)=\sqrt{-x}$ horizontally right by 1 unit to get the graph of $f\left(x\right)=\sqrt{1-x}$ .

  

Intercepts are the points where the graph touches or crosses the coordinate axes. In $f\left(x\right)=\sqrt{1-x}$ , the graph touches the coordinate axes at $\left(1,0\right)$ and $\left(0,1\right)$

So, the intercepts are $\left(1,0\right),\left(0,1\right)$ .

The points on the graph have $x$ coordinates between $-\infty$ and $1$ , and for each number $x$ between $-\infty$ and 1, there is a point $\left(x,f\left(x\right)\right)$ on the graph. So, the domain is $\left\{x|x\le 1\right\}$ or the interval $\left[-\infty ,1\right]$ .

The points on the graph have $y$ coordinates between $0$ and $\infty$ , and for each number $y$ , there is at least one number $x$ in the domain.

So, the range of $F\left(x\right)$ is $\left\{y|y\ge 0\right\}$ or the interval $\left[0,\infty \right)$ .

Therefore, the intercepts are $\left(1,0\right),\left(0,1\right)$ , domain is $\left\{x|x\le 1\right\}$ or the interval $\left[-\infty ,1\right]$ , and range is $\left\{y|y\ge 0\right\}$ or the interval $\left[0,\infty \right)$ .