# Graph of Inverse Function Worksheet

Graph of Inverse Function Worksheet
• Page 1
1.
The weight ($w$) of the soccer ball used by the age group 8 to 12, vary from 373 grams to 403 grams, inclusive. Write an absolute value inequality describing the weights of soccer balls that are outside the acceptable range.
 a. | $w$ - 403 | < 373 b. | $w$ - 388 | < 15 c. | $w$ - 388 | > 15 d. $w$ < 373 or $w$ > 403

#### Solution:

w be the weight of a soccer ball, then 373 ≤ w ≤ 403
[Weight of a soccer ball vary from 373 grams to 403 grams, inclusive.]

w < 373 or w > 403
[Outside the acceptable range.]

w < 388 - 15 or w > 388 + 15
[Average of 373 and 403 is 388.]

w - 388 < - 15 or w - 388 > 15
[Rearrange the terms.]

- (w - 388) > 15 or (w - 388) > 15
[If a < - b then - a > b.]

| w - 388 | > 15
[Use: if - x > a or x > a, then | x | > a.]

2.
Earth is not a perfect sphere, but nearly so, with the approximate surface area 510926783 km2. Find the earth's diameter approximately. (Use $\pi$ = 3.14)
 a. 7198 km b. 12756 km c. 6378 km d. 3189 km

#### Solution:

Surface area of a sphere, S = 4πr2, where r is the radius of the sphere.

d = 2r = 2s4π = sπ

d = 5109267833.14 12756 km.

So, the diameter of the Earth approximately is 12756 km.

3.
The diameter of a tennis ball is between 6.34 cm and 6.68 cm. Write an absolute value inequality that describes the diameter ($d$) range of a tennis ball.
 a. 6.34 < $d$ < 6.68 b. | $d$ - 6.51 | > 0.17 c. | $d$ - 6.51 | < 0.17 d. | $d$ - 6.68 | < 6.34

#### Solution:

Let d be the diameter of a tennis ball, then 6.34 < d < 6.68
[Diameter of a tennis ball vary from 6.34 cm to 6.68 cm.]

6.34 - 6.51 < d - 6.51 < 6.68 - 6.51
[Average of 6.34 and 6.68 is 6.51.]

- 0.17 < d - 6.51 < 0.17
[Simplify.]

| d - 6.51 | < 0.17
[If - a < ( x - k ) < a, then | x - k | < a.]

4.
A rectangular gift box with volume (V) 968 cm3 and a height of $h$ cm has a square base. Write an equation to find the length of one side of the base and find the value of it for $h$ = 8 cm.
 a. $\sqrt{\frac{V}{h}}$ ; 11 cm b. $\sqrt{Vh}$ ; 88 cm c. $\frac{\sqrt{V}}{h}$ ; 6.2 cm d. $\sqrt{\frac{V}{h}}$ ; 121 cm

#### Solution:

Volume of the rectangular gift box, V = base area × height

Let x be the base length of a rectangular gift box.

V = x2h x2 = Vh
[Base area = x × x = x2.]

So, x = Vh

x = 9688 = 11
[Replace V = 968 cm3 and h = 8 cm.]

So, the length of the one side of the base of a rectangular gift box is 11 cm.

5.
The manufacturer of a particular equipment for fitness, claims that the reduction in weight ($w$) resulting by using this equipment satisfies the inequality | $w$ - 18 | ≤ 10. Find the maximum weight in pounds reduced by using that equipment as claimed by the manufacturer.
 a. 18 pounds b. 28 pounds c. 10 pounds d. 180 pounds

#### Solution:

| w - 18 | ≤ 10
[Weight reduction inequality.]

- 10 ≤ (w - 18) ≤ 10
[If | x | ≤ a then - axa.]

- 10 + 18 ≤ w ≤ 10 + 18
[Add 18 to each term of the inequality.]

8 ≤ w ≤ 28
[Simplify.]

So, the maximum weight reduced by using the equipment is 28 pounds.

6.
Find the equation of the inverse of $y$ = 2$x$.
 a. $y$ = - 2$x$ b. $y$ = $\frac{1}{2}$$x$ c. $x$ = $\frac{1}{2}$$y$ d. $y$ = - $\frac{1}{2}$$x$

#### Solution:

y = 2x

x = 2y
[Interchange x and y.]

y = x2
[Solve for y.]

So, the equation of the inverse is y = 1 / 2x.

7.
Find the equation of the inverse of $y$ = $x$ - 2.
 a. $x$ = $y$ + 2 b. $x$ = $\sqrt{y}$ - 2 c. $y$ = $x$ - 2 d. $y$ = $x$ + 2

#### Solution:

y = x - 2

x = y - 2
[Interchange x and y.]

y = x + 2
[Solve for y.]

So, the equation of the inverse is y = x + 2.

8.
Find the equation of the inverse of $f$($x$) = .
[Hint: Replace $f$($x$) with $y$]
 a. $y$ = - 2$x$ + 3 b. $y$ = 3$x$ - 2 c. $x$ = d. Inverse does not exist

#### Solution:

f(x) = x + 23

y = x + 23
[Replace f(x) by y.]

x = y + 23
[Interchange x and y.]

y = 3x - 2
[Solve for y.]

So, the equation of the inverse is y = 3x - 2.

9.
Find the equation of the inverse of $y$ = $\frac{1}{7}$ $x$2.
 a. $y$ = ± $\sqrt{7x}$ b. $x$ = $\frac{1}{7}$$y$ c. $y$ = - $\sqrt{7}$$x$ d. $y$ = $\sqrt{7}$$x$

#### Solution:

y = 1 / 7x2

x = 1 / 7y2
[Interchange x and y.]

7x = y2
[Multiply throughout by 7.]

y = ± 7x
[Solve for y.]

So, the equation of the inverse is y = ± 7x.

10.
Find the equation of the inverse of $y$ = $x$2 - 6.
 a. $y$ = ± b. $y$ = - only c. $y$ = only d. $x$2 = $y$² - 6

#### Solution:

y = x² - 6

x = y² - 6
[Interchange x and y.]

x + 6 = y²
[Add 6 to both sides of the equation.]

y = ± x + 6
[Solve for y.]

So, the equation of the inverse is y = ± x + 6.