A function f : [- 7, 6) → R is defined as follows f(x) = {(x^2 + 2x + 1; -7 ≤ x < -5), (x + 5; -5 ≤ x ≤ 2), (x - 1; 2 < x < 6)

Question

A function f : [- 7, 6) → R is defined as follows \(f(x) = \begin{cases} x^2+2x+1; & \quad \text{} -7≤x<-5 \text{ }\\ x+5; & \quad \text{} -5≤x≤2 \text{ }\\x-1; & \quad \text{} 2<x<6 \end{cases}\)

(i) 2f(-4) + 3f(2),

(ii) f(-7) - f(-3)

(iii) (4f(-3) + 2f(4))/(f(-6) - 3f(1))

Answer 1

Given, f(x) = x² + 2x + 1; x = {-7, -6}

f(x) = x + 5; x = {-5, -4, -3, -2, -1, 0, 1, 2}

f(x) = x – 1; x{3, 4, 5}

(i) 2f(- 4) + 3f(2)

f(x) = x + 5

f(-4) = -4 + 5 = 1

f(2) = 2 + 5 = 7

∴ 2f(-4) + 3 f(2) = 2(1) + 3(7) = 2 + 21 = 23

(ii) f(-7) – f(-3)

f(x) = x² + 2x + 1

f(-7) = (-7)² + 2(-7) + 1 = 49 – 14 + 1 = 36

f(x) = x + 5

f(-3) = -3 + 5 = 2

∴ f(-7) – f(-3) = 36 – 2 = 34

(iii) (4f(-3) + 2f(4))/(f(-6) - 3f(1))

f(x) = x + 5

f(-3) = -3 + 5 = 2

f(x) = x – 1

f(4) = 4 – 1 = 3

f(x) = x² + 2x + 1

f(-6) = (-6)² + 2(-6) + 1 = 36 – 12 + 1 = 25

f(x) = x + 5

f(1) = 1 + 5 = 6