Exercise 1.1.59 (59-64)

Remark for Exercises 59, 60, 63, 64. Line segment is a graph of an affine function $f:[a,b]\to ℝ$, i.e., of a function given by the formula

where $m\in ℝ,b\in ℝ$ are some constants. For any affine function $f$ this formula is true:

where $(x_1,y_1),(x_2,y_2)$ are given two points that lie on the graph of the function.

59.

In this exercise, (1 ) translates to: $$\frac{x-1}{5-1}=\frac{L(x)-(-3)}{7-(-3)}$$

$$\frac{x-1}{4}=\frac{L(x)+3}{10}$$

$$4L(x)+12=10x-10$$

$$4L(x)=10x-22$$

$$L(x)=2.5x-5.5.$$

This line segment is defined by the graph of the function:

$$L:[1,5]\to ℝ,L(x)=2.5x-5.5..$$

60.

In this exercise, (1 ) translates to: $$\frac{x+5}{7-(-5)}=\frac{L(x)-10}{-10-10}$$

$$\frac{x+5}{12}=\frac{L(x)-10}{-20}$$

$$-20x-100=12L(x)-120$$

$$-12L(x)=20x-20$$

$$L(x)=-1\frac{2}{3}x+1\frac{2}{3}.$$

This line segment is thus defined by the graph of the function:

$$L:[-5,7]\to ℝ,L(x)=-1\frac{2}{3}x+1\frac{2}{3}.$$

61.

We must first solve this equation for $y$: $${(y-1)}^2=-x$$

$$y-1=\pm \sqrt{-x}$$

$$y=\pm \sqrt{-x}+1.$$

To get the bottom half of this function, the formula must look like this:

$$y=-\sqrt{-x}+1.$$

The bottom half of the parabola is thus defined by the graph of the function:

$$y:(-\infty ,0]\to ℝ,y=-\sqrt{-x}+1.$$

62.

We must first solve this equation for $y$: $${(y-2)}^2=4-x^2$$

$$y-2=\pm \sqrt{4-x^2}$$

$$y=\pm \sqrt{4-x^2}+2$$

To get the top half of the circle, the formula must look like this:

$$y=\sqrt{4-x^2}+2.$$

The top half of the circle is thus defined by the graph of the function:

$$y:[-2,2]\to ℝ,y=-\sqrt{4-x^2}+2.$$

63.

The graph in the picture is a piecewise-defined function and is defined by two linear functions. To find out these functions, we must take two points on each part:

1.

Here, (1 ) translates to: $$\frac{x-0}{3-0}=\frac{g(x)-3}{0-3}$$

$$\frac{x}{3}=\frac{g(x)-3}{-3}$$

$$-3x=3g(x)-9(÷-3)$$

$$x=-g(x)+3$$

$$g(x)=-x+3.$$

2.

Here, (1 ) translates to: $$\frac{x-4}{5-4}=\frac{K(x)-2}{4-2}$$

$$\frac{x-4}{1}=\frac{K(x)-2}{2}$$

$$x-4=\frac{K(x)-2}{2}$$

$$K(x)-2=2x-8$$

$$K(x)=2x-6.$$

Thus, the function $h$ we’re looking for is:

$$h:[0,5]\to ℝ,h(x)=\{\begin{array}{cc}-x+3,\hfill & \text{if }0\le x\le 3\hfill \\ 2x-6,\hfill & \text{if }3<x\le 5\hfill \end{array}$$

64.

The function we see on the graph consists of three parts:

• (1st part) 1st part of the function is linear. Thus, to find the formula of this linear function $G$ we have to find two points that belong to the graph: $(-4,3)$ and $(-2,0)$. Therefore, (1 ) translates to:

  $$\frac{x+4}{-2+4}=\frac{G(x)-3}{0-3}$$
  $$\frac{x+4}{2}=\frac{G(x)-3}{-3}$$
  $$-3x-12=2G(x)-6$$
  $$-2G(x)=3x+6$$
  $$G(x)=-1.5x-3.$$

• (2nd part) 2nd part of the function looks like top half of the circle with radius in 2 unit steps. That is why, the formula of the second part is:

  $$K(x)=\sqrt{4-x^2}.$$

• (3rd part) 3rd part of the function is also linear. Thus, to find the formula of this linear function $Q$ we have to find two points that belong to the graph: $(4,3)$ and $(2,0)$. Therefore, (1 ) translates to:

  $$\frac{x-4}{2-4}=\frac{Q(x)-3}{0-3}$$
  $$\frac{x-4}{-2}=\frac{Q(x)-3}{-3}$$
  $$-3x+12=-2Q(x)+6$$
  $$2Q(x)=3x-6$$
  $$Q(x)=1.5x-3.$$

Thus, we define the function $q$ we’re looking for as follows:

$$q:[-4,4]\to ℝ,q(x)=\{\begin{array}{cc}-1.5x-3,\hfill & \text{if }x\le -2\hfill \\ \sqrt{4-x^2},\hfill & \text{if }-2<x<2\hfill \\ 1.5x-3,\hfill & \text{if }x\ge 2\hfill \end{array}$$