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.
1. Answer :
(i) f(2) : f(x) = 5x - 7 f(2) = 5(2) - 7 = 10 - 7 = 3 (iii) f(-1) : f(x) = 5x - 7 f(-1) = 5(-1) - 7 = -5 - 7 = -12 |
(ii) f(0) : f(x) = 5x - 7 f(0) = 5(0) - 7 = 0 - 7 = -7 (iv) f(½) : f(x) = 5x - 7 f(½) = 5(½) - 7 = ⁵⁄₂ - ¹⁴⁄₂ = ⁽⁵ ⁻ ¹⁴⁾⁄₂ = ⁻¹¹⁄₂ |
2. Answer :
(i) g(0) : g(y) = -5y + 2 g(0) = -5(0) + 2 = 0 + 2 = 2 (iii) g(-4) : g(y) = -5y + 2 g(-4) = -5(-4) + 2 = 20 + 2 = 22 |
(ii) g(5) : g(y) = -5y + 2 g(3) = -5(3) + 2 = -15 + 2 = -13 (iv) g(⁻¹⁄₄) : g(y) = -5y + 2 g(⁻¹⁄₄) = -5(⁻¹⁄₄) + 2 = ⁵⁄₄ + 2 = ⁵⁄₄ + ⁸⁄₄ = ⁽⁵ ⁺ ⁸⁾⁄₄ = ¹³⁄₄ |
3. Answer :
(i) h(-2) : h(z) = 5 - (½)z h(-2) = 5 - (½)(-2) = 5 + 1 = 6 (iii) h(0) : h(z) = 5 - (½)z h(0) = 5 - (½)(0) = 5 - 0 = 5 |
(ii) h(4) : h(z) = 5 - (½)z h(4) = 5 - 2 = 5 - 2 = 3 (iv) h(0.5) : h(z) = 5 - (½)z h(0.5) = 5 - (½)(0.5) = 5 - (½)(½) = 5 - ¼ = ²⁰⁄₄ - ¼ = ⁽²⁰ ⁻ ¹⁾⁄₄ = ¹⁹⁄₄ |
4. Answer :
(i) f(x) = 8 : f(x) = 8 3x + 2 = 8 3x = 6 x = 3 (iii) f(x) = 0.5 : f(x) = 0.5 3x + 2 = 0.5 2x = -1.5 2x = ⁻³⁄₂ x = ⁻³⁄₄ |
(ii) f(x) = -7 : f(x) = -7 3x + 2 = -7 3x = -9 x = -3 (iv) f(x) = ⁻¹⁄₄ : f(x) = ⁻¹⁄₄ 3x + 2 = ⁻¹⁄₄ 3x = ⁻¹⁄₄ - 2 3x = ⁻¹⁄₄ - ⁸⁄₄ 3x = ⁽⁻¹ ⁻ ⁸⁾⁄₄ 3x = ⁻⁹⁄₄ x = ⁻³⁄₄ |
5. Answer :
(i) g(y) = 26 : g(y) = 26 -5y + 1 = 26 -5y = 25 5y = -25 y = -5 (iii) g(y) = -14 : g(y) = -14 -5y + 1 = -14 -5y = -15 5y = 15 y = 3 |
(ii) g(y) = 0 : g(y) = 0 -5y + 1 = 0 -5y = -1 5y = 1 y = 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 = ⁻⁷⁄₂ 5y = ⁷⁄₂ y = ⁷⁄₁₀ |
6. Answer :
(i) h(z) = 7 : h(z) = 7 (⁻¹⁄₂)z + 7 = 7 (⁻¹⁄₂)z = 0 -z = 0 z = 0 (iii) h(z) = 0 : h(z) = 0 (⁻¹⁄₂)z + 7 = 0 (⁻¹⁄₂)z = -7 -z = -14 z = 14 |
(ii) h(z) = -2 : h(z) = -2 (⁻¹⁄₂)z + 7 = -2 (⁻¹⁄₂)z = -9 -z = -18 z = 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 = ⁻²⁷⁄₂ z = ²⁷⁄₂ |
7. Answer :
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
8. Answer :
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
9. Answer :
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.
10. Answer :
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.
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