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A function f: [-5, 9] → R is defined as follows

\(f(n) = \begin{cases} 6x+1 & \quad \text{if } -5≤x<2 \text{ }\\ 5x^2-1 & \quad \text{if } 2≤x<6 \text{ }\\3x-4& \quad \text{if }6≤x≤9 \end{cases}\)

Find 

(i) f(-3) + f(2)  

(ii) f(7) - f(1)

(iii) 2f(4) + f(8) 

(iv) (2f(-2) - f(6))/(f(4) + f(-2))

1 Answer

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Best answer

f(x) = 6x + 1 ; x = {-5,-4,-3,-2,-1,0,1} 

f(x) = 5x2 – 1 ; x = {2, 3, 4, 5} 

f(x) = 3x – 4 ; x = {6, 7, 8, 9}

(i) f(-3) + f(2) 

f(x) = 6x + 1 

f(-3) = 6(-3) + 1 = -18 + 1 = -17 

f(x) = 5x2 – 1 

f(2) = 5(2)2 – 1 = 20 – 1 = + 19 

f(-3) + f(2) = – 17 + 19 = 2

(ii) f(7) – f(1) 

f(x) = 3x – 4 

f(7) = 3(7) – 4 = 21 – 4 = 17 

f(x) = 6x + 1 

f(1) = 6(1) + 1 = 6 + 1 = 7 

f(7) – f(1) = 17 – 7 = 10

(iii) 2f(4) + f(8) 

f(x) = 5x2 – 1 

f(4) = 5(4)2 – 1 = 5(16) – 1 

= 80 – 1 = 79 

f(x) = 3x – 4 

f(8) = 3(8) – 4 = 24 – 4 = 20

2f(4) + f(8) = 2(79) + 20 

= 158 + 20 = 178

(iv) (2f(-2) - f(6))/(f(4) + f(-2))

f(x) = 6x + 1 

f(-2) = 6(-2) + 1 = -12 + 1 = -11 

f(x) = 3x – 4 

f(6) = 3(6) – 4 = 18 – 4 = 14 

f(x) = 5x2 – 1 

f(4) = 5(4)2 – 1 = 5(16) – 1 

= 80 – 1 = 79 

f(x) = 6x + 1

f(-2) = 6(-2) + 1 = -12 + 1 = -11

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