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)

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 + 2f(-1) = 3(-1) + 2f(-1) = -3 + 2f(-1) = -1 f(x) = 3x + 2f(-2) = 3(-2) + 2f(-2) = -6 + 2f(-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?.

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.

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).

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)?

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).

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).

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)?

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

f(x) = ax3 + b

 f(-1) = a(-1)3 + b4 = a(-1) + b4 = -a + b-a + b = 4 ----(1) f(1) = a(1)3 + b10 = a(1) + b10 = a + ba + b = 10 ----(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)?

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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