# FUNCTION NOTATION INPUT AND OUTPUT WORKSHEET

Problem 1 :

If f(x) = 5x - 7, evaluate f(x) in each case for the given value of x.

(i) f(2)

(ii) f(0)

(iii) f(-1)

(iv) f(½)

Problem 2 :

If g(y) = -5y + 2, evaluate g(y) in each case for the given values of y.

(i) g(0)

(ii) g(5)

(iii) g(-4)

(iv) g(⁻¹⁄₄)

Problem 3 :

If h(z) = 5 - (½)z, evaluate h(z) in each case for the given value of z.

(i) h(-2)

(ii) h(4)

(iii) h(0)

(iv) h(0.5)

Problem 4 :

If f(x) = 3x + 2, find the value of x in each case for the given value of f(x).

(i) f(x) = 8

(ii) f(x) = -7

(iii) f(x) = 0.5

(iv) f(x) = ⁻¹⁄₄

Problem 5 :

If g(y) = -5y + 1, find the value of y in each case for the given value of g(y).

(i) g(y) = 26

(ii) g(y) = 0

(iii) g(y) = -14

(iv) g(y) = ⁻⁵⁄₂

Problem 6 :

If h(z) = (⁻¹⁄₂)z + 7, find the value of z in each case for the given value of h(z).

(i) h(z) = 7

(ii) h(z) = -2

(iii) h(z) = 0

(iv) h(z) = ¹⁄₄

Problem 7 :

g(y) = 7y - a

In the function above, a is a constant and g(5) = 17. What is the value of g(-2)?

Problem 8 :

If h(1 - z) = 3z - 2, find the value of h(-1).

Problem 9 :

An online film renting service charges a one time fee of \$35 for 1 year subscription and \$7 for each film rented in that year. Find the function which models the total charge for x number of films in a year. And also, find the total charge for 15 films rented in a year.

Problem 10 :

A tank contains 27.5 gallons of water now and water is being pumped into the tank at a rate of 2 gallons per minute. Write the function that models the amount of water in the tank in x minutes from now. And also, find the amount of water in the tank 20 minutes from now.

 (i) f(2) :f(x) = 5x - 7f(2) = 5(2) - 7= 10 - 7= 3 (iii) f(-1) :f(x) = 5x - 7f(-1) = 5(-1) - 7= -5 - 7= -12 (ii) f(0) :f(x) = 5x - 7f(0) = 5(0) - 7= 0 - 7= -7 (iv) f(½) :f(x) = 5x - 7f(½) = 5(½) - 7= ⁵⁄₂ - ¹⁴⁄₂= ⁽⁵ ⁻ ¹⁴⁾⁄₂= ⁻¹¹⁄₂

 (i) g(0) :g(y) = -5y + 2g(0) = -5(0) + 2= 0 + 2= 2 (iii) g(-4) :g(y) = -5y + 2g(-4) = -5(-4) + 2= 20 + 2= 22 (ii) g(5) :g(y) = -5y + 2g(3) = -5(3) + 2= -15 + 2= -13 (iv) g(⁻¹⁄₄) :g(y) = -5y + 2g(⁻¹⁄₄) = -5(⁻¹⁄₄) + 2= ⁵⁄₄ + 2= ⁵⁄₄ + ⁸⁄₄= ⁽⁵ ⁺ ⁸⁾⁄₄= ¹³⁄₄

 (i) h(-2) :h(z) = 5 - (½)zh(-2) = 5 - (½)(-2)= 5 + 1= 6 (iii) h(0) :h(z) = 5 - (½)zh(0) = 5 - (½)(0)= 5 - 0= 5 (ii) h(4) :h(z) = 5 - (½)zh(4) = 5 - 2= 5 - 2= 3 (iv) h(0.5) :h(z) = 5 - (½)zh(0.5) = 5 - (½)(0.5)= 5 - (½)(½)= 5 - ¼= ²⁰⁄₄ - ¼= ⁽²⁰ ⁻ ¹⁾⁄₄= ¹⁹⁄₄

 (i) f(x) = 8 :f(x) = 83x + 2 = 83x = 6x = 3 (iii) f(x) = 0.5 :f(x) = 0.53x + 2 = 0.52x = -1.52x = ⁻³⁄₂x = ⁻³⁄₄ (ii) f(x) = -7 :f(x) = -73x + 2 = -73x = -9x = -3 (iv) f(x) = ⁻¹⁄₄ :f(x) = ⁻¹⁄₄3x + 2 = ⁻¹⁄₄3x = ⁻¹⁄₄ - 23x = ⁻¹⁄₄ - ⁸⁄₄3x = ⁽⁻¹ ⁻ ⁸⁾⁄₄3x = ⁻⁹⁄₄x = ⁻³⁄₄

 (i) g(y) = 26 :g(y) = 26-5y + 1 = 26-5y = 255y = -25y = -5 (iii) g(y) = -14 :g(y) = -14-5y + 1 = -14-5y = -155y = 15y = 3 (ii) g(y) = 0 :g(y) = 0-5y + 1 = 0-5y = -15y = 1y = ⅕ (iv) g(y) = ⁻⁵⁄₂ :g(y) = ⁻⁵⁄₂-5y + 1 = ⁻⁵⁄₂-5y = ⁻⁵⁄₂ - 1-5y = ⁻⁵⁄₂ - ²⁄₂-5y = ⁽⁻⁵ ⁻ ²⁾⁄₂-5y = ⁻⁷⁄₂5y = ⁷⁄₂y = ⁷⁄₁₀

 (i) h(z) = 7 :h(z) = 7(⁻¹⁄₂)z + 7 = 7(⁻¹⁄₂)z = 0-z = 0z = 0 (iii) h(z) = 0 :h(z) = 0(⁻¹⁄₂)z + 7 = 0(⁻¹⁄₂)z = -7-z = -14z = 14 (ii) h(z) = -2 :h(z) = -2(⁻¹⁄₂)z + 7 = -2(⁻¹⁄₂)z = -9-z = -18z = 18 (iv) h(z) = ¹⁄₄ :h(z) = ¹⁄₄(⁻¹⁄₂)z + 7 = ¹⁄₄(⁻¹⁄₂)z = ¹⁄₄ - 7(⁻¹⁄₂)z = ¹⁄₄ - ²⁸⁄₄(⁻¹⁄₂)z = ⁽¹ ⁻ ²⁸⁾⁄₄(⁻¹⁄₂)z = ⁻²⁷⁄₄-z = ⁻²⁷⁄₂z = ²⁷⁄₂

To find the value of g(-2), first we have to find the value of the constant a in the given function g(y) = 7y - a.

g(y) = 7y - a

g(5) = 7(5) - a

g(5) = 35 - a ----(1)

It is given that

g(5) = 17 ----(2)

From (1) and (2),

35 - a = 17

-a = -18

a = 18

Now, the given function is

g(y) = 7y - 18

Substitute y = -2.

g(-2) = 7(-2) - 18

= -14 - 18

= -32

The given function is not in the form h(z). So, we can not substitute z = -1 in h(1 - z) First, we have to change the given function to the regular form h(x). Then, we have to substitute x = -1 to evaluate h(-1).

h(1 - z) = 3z - 2 ----(1)

Let x = 1 - z.

Solve for z in terms of x.

x = 1 - z

x + z = 1

z = 1 - x

Substitute x for '1 - z' and '1 - x' for z in (1).

h(x) = 3(1 - x) - 2

h(x) = 3 - 3x - 2

h(x) = -3x + 1

Substitute x = -1.

h(-1) = -3(-1) + 1

h(-1) = 3 + 1

h(-1) = 4

It is given that that charge for each film rented is \$12.

1 film ----> \$7

x(1 film) ----> x(\$7)

x films = 7x

Total charge for 1 year (including onetime fee) :

= 7x + 50

Therefore, the function which models the total charge for 1 year :

T(x) = 7x + 50

To find the total charge for 15 films, substitute x = 15 into the above function.

T(15) = 7(15) + 35

= 105 + 35

= 140

The total charge for 15 films rented in a year is \$140.

It is given that water is being pumped into the tank at a rate 2 gallons per minute.

1 minute ----> 2 gallons

x(1 minute) ----> x(2 gallons)

x minutes ----> 2x gallons

Total amount of water in the tank in x minutes from now (including the initial 27.5 gallons of water) :

= 2x + 27.5

Therefore, the function which models the amount of water in the tank in x minutes from now :

W(x) = 2x + 27.5

To find the amount of water in 20 minutes from now, substitute x = 20 into the above function.

W(40) = 2(40) + 27.5

= 80 + 27.5

= 107.5

The amount of water in the tank in x minutes from now is 107.5 gallons.

## Practice Questions

Kindly mail your feedback to v4formath@gmail.com

## Recent Articles

1. ### Cubes and Cube Roots

Dec 11, 23 08:32 AM

Cubes and Cube Roots - Concepts - Examples

2. ### Worksheet on Speed Distance and Time

Dec 10, 23 10:09 PM

Worksheet on Speed Distance and Time