Chapter 1, Problem 18RE

To determine

To find: An expression function for the graph which satisfies the given conditions.

Answer to Problem 18RE

Solution:

The equation of the function is $f\left(x\right)=\{\begin{array}{l}-2x-2\text{ on }\left[-2,-1\right)\\ \sqrt{1-x^2}\text{ on }\left[-1,-1\right]\end{array}$.

Explanation of Solution

Given:

The graph has a line segment connecting (−2, 2) and (−1, 0) and it consist a top half of the circle with centre (0, 0) and radius 1.

Calculation:

Find the slope of the line segment joining the points (−2, 2) and (−1, 0) as follows.

$m=\frac{y_2-y_1}{x_2-x_1}$

$\begin{array}{c}m=\frac{0-2}{-1-\left(-2\right)}\\ =\frac{-2}{1}\\ =-2\end{array}$

Thus, the slope of the line segment is $m=-2$.

Find the y-intercept of the line segment joining the points (−2, 2) and (−1, 0) as follows.

$\begin{array}{l}y=mx+c\\ 0=\left(-2\right)\left(-1\right)+c\left[\because m=-2\right]\\ 0=2+c\\ c=-2\end{array}$

Thus, y-intercept is $c=-2$.

Therefore, the equation of the line segment is, $y=-2x-2$ where $-2\le x<-1$ (1)

Also there exist top half of the circle with centre (0, 0) and radius 1.

The standard equation of the circle which passes through the (0, 0) and radius 1 is $x^2+y^2=1$.

Solve the equation for y as follows.

$\begin{array}{l}{y}^2=1-x^2\\ y=\pm \sqrt{1-x^2}\end{array}$

Since the graph consist only top half of the circle, consider only the positive root. That is $y=\sqrt{1-x^2}$.

Therefore, equation of the top half of the circle is $y=\sqrt{1-x^2},-1\le x\le 1$ (2)

Combine the equations (1) and (2), the function of the graph is $f\left(x\right)=\{\begin{array}{l}-2x-2\text{ on }\left[-2,-1\right)\\ \sqrt{1-x^2}\text{ on }\left[-1,-1\right]\end{array}$.