Question 1 :
If f(x) = 3x + 2, find each value.
(i) f(-2)
(ii) f(k - 2)
(iii) -2[f(-1)] + f(-2)
(iv) f(-ˣ⁄₂ + 1)
Answer :
Part (i) :
f(x) = 3x + 2
f(-2) = 3(-2) + 2
f(-2) = -6 + 2
f(-2) = -4
Part (ii) :
f(x) = 3x + 2
f(k - 2) = 3(k - 2) + 2
f(k - 2) = 3k - 6 + 2
f(k - 2) = 3k - 4
Part (iii) :
f(x) = 3x + 2 f(-1) = 3(-1) + 2 f(-1) = -3 + 2 f(-1) = -1 |
f(x) = 3x + 2 f(-2) = 3(-2) + 2 f(-2) = -6 + 2 f(-2) = -4 |
-2[f(-1)] + f(-2) :
= -2(-1) + (-4)
= 2 - 4
= -2
Part (iv) :
f(x) = 3x + 2
f(-ˣ⁄₂ + 2) = 3(-ˣ⁄₂ + 2) + 2
f(-ˣ⁄₂ + 2) = -³ˣ⁄₂ + 6 + 2
f(-ˣ⁄₂ + 2) = -³ˣ⁄₂ + 8
Question 2 :
If f(x) = -2x + 7, what is f(ˣ⁄₂ + 3) equal to?.
Answer :
f(x) = -2x + 7
f(ˣ⁄₂ + 3) = -2(ˣ⁄₂ + 3) + 7
f(ˣ⁄₂ + 3) = -x - 6 + 7
f(ˣ⁄₂ + 3) = -x + 1
Question 3 :
If f(x) = ax + 3 and f(2) = 13, find the value of a.
Answer :
f(x) = ax + 3
Given f(2) = 13.
f(2) = 13
a(2) + 3 = 13
2a + 3 = 13
2a = 10
a = 5
Question 4 :
f(x) = kx3 + 3
For the function f(x) defined above, k is a constant and f(-1) = 5. What is the value of f(1).
Answer :
f(x) = kx3 + 3
Given f(-1) = 5.
f(-1) = 5
k(-1)3 + 3 = 5
k(-1) + 3 = 5
-k + 3 = 5
-k = 2
k = -2
Then, we have
f(x) = -2x3 + 3
f(1) = -2(1)3 + 3
f(1) = -2(1) + 3
f(1) = -2 + 3
f(1) = 1
Question 5 :
f(x) = x2 - b
In the function above, b is a constant. If f(-2) = 7, what is the value of f(b)?
Answer :
f(x) = x2 - b
Given f(-2) = 7.
f(-2) = 7
(-2)2 - b = 7
4 - b = 7
-b = 3
b = -3
f(x) = x2 - (-3)
f(x) = x2 + 3
f(b) = f(-3)
f(b) = (-3)2 + 3
f(b) = 9 + 3
f(b) = 12
Question 6 :
If f(1 - x) = 5x + 2, find f(y).
Answer :
f(1 - x) = 5x + 2 ----(1)
Let y = 1 - x. Solve for x in terms of y.
x = 1 - y
Substitute '1 - x = y' and 'x = 1 - y' in (1).
f(y) = 5(1 - y) + 2
f(y) = 5 - 5y + 2
f(y) = -5y + 7
Question 7 :
If f(x + 2) = 3x - 5, find f(x).
Answer :
f(x + 2) = 3x - 5 ----(1)
Let y = x + 2. Solve for x in terms of y.
x = y - 2
Substitute 'x + 2 = y' and 'x = y - 2' in (1).
f(y) = 3(y - 2) - 5
f(y) = 3y - 6 - 5
f(y) = 3y - 11
Replace y by x.
f(x) = 3x - 11
Question 8 :
If f(x + 1) = -ˣ⁄₂ + 6, what is the value of f(-3)?
Answer :
f(x + 1) = -ˣ⁄₂ + 6 ----(1)
Let y = x + 1. Solve for x in terms of y.
y = x + 1
Substitute 'x + 1 = y' and 'x = y - 1' in (1).
f(y) = -⁽ʸ ⁻ ¹⁾⁄₂ + 6
Substitute y = -3.
f(-3) = -⁽⁻³ ⁻ ¹⁾⁄₂ + 6
f(-3) = -⁽⁻⁴⁾⁄₂ + 6
f(-3) = -(-2) + 6
f(-3) = 2 + 6
f(-3) = 8
Question 9 :
f(x) = ax3 + b
In the function defined above, a and b are constants. If f(-1) = 4 and f(1) = 10, find the value of b.
Answer :
f(x) = ax3 + b
f(-1) = a(-1)3 + b 4 = a(-1) + b 4 = -a + b -a + b = 4 ----(1) |
f(1) = a(1)3 + b 10 = a(1) + b 10 = a + b a + b = 10 ----(2) |
Add (1) and (2).
2b = 14
b = 7
Question 10 :
f(x) = x3 + 2
g(x) = 2x
The functions f and g are defined above. If f(b) = 29, what is the value of g(b)?
Answer :
f(b) = 29
b3 + 2 = 29
b3 = 27
b3 = 33
b = 3
g(b) = g(3)
g(b) = 2(3)
g(b) = 6
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