# EVAlUATE COMPOSITION OF FUNCTIONS FOR THE GIVEN VALUE

The composition of a function g with a function f is :

h(x) = g(f (x))

The domain of h is the set of all x-values such that x is in the domain of f and f (x) is in the domain of g.

Let f(x) = 2x – 1, g(x) = 3x, and h(x) = x2 + 1.Compute the following :

Problem 1 :

f(g(-3))

Solution :

f(x) = 2x – 1 and g(x) = 3x

 Apply x = -3 in g(x)g(-3) = 3(-3)g(-3) = -9f(g(-3)) = f(-9) Apply x = -9 in f(x)f(-9) = 2(-9) – 1= -18 – 1f(-9) = -19

f(g(-3)) = -19

Problem 2 :

f(h(7))

Solution :

f(x) = 2x – 1

h(x) = x2 + 1

 f(h(7)) = ?Apply x = 7 in h(x)h(7) = (7)2 + 1h(7) = 49 + 1h(7) = 50 f(h(7)) = f(50)Apply x = 50 in h(x).= 2(50) – 1= 100 – 1f(h(7)) = 99

f(h(7)) = 99

Problem 3 :

(g  h)(24)

Solution :

g(x) = 3x

h(x) = x2 + 1

(g  h)(24) = g(h(24))

 Apply x = 24 in h(x)h(24) = (24)2 + 1= 577 g(h(24)) = g(577)Apply x = 577 in g(x)= 3(577)= 1731

(g  h)(24) = 1731

Problem 4 :

f(g(h(2)))

Solution :

f(x) = 2x – 1

g(x) = 3x

h(x) = x2 + 1 and f(g(h(2))) = ?

 Apply x = 2 in h(x)h(2) = (2)2 + 1h(2) = 5 g(h(2)) = g(5)Now apply x = 5 in g(5) = 3(5)g(5) = 15

f(g(h(2))) = f(15)

Apply x = 15 in f(x).

= 2(15) – 1

f(g(h(2))) = 29

Problem 5 :

h(g(f(5)))

Solution :

Given functions are f(x) = 2x – 1, g(x) = 3x and h(x) = x2 + 1

h(g(f(5))) = ?

 Apply x = 5 in f(x)f(5) = 2(5) – 1f(5) = 9 g(f(5)) = g(9)Apply g(9) in g(x)= 3(9)g(9) = 27

h(g(f(5))) = h(27)

= (27)2 + 1

= 730

h(g(f(5))) = 730

Problem 6 :

g(f(h(-6)))

Solution :

Given functions are (x) = 2x – 1, g(x) = 3x and h(x) = x2 + 1

g(f(h(-6))) = ?

 Apply x = -6 in h(x)h(-6) = (-6)2 + 1h(-6) = 37f(h(-6)) = f(37) Apply x = 37 in f(x). = 2(37) – 1= 74 – 1= 73

g(f(h(-6))) = g(73)

= 3(73)

= 219

g(f(h(-6))) = 219

Problem 7 :

If f(x) = 3x - 5 and g(x) = x2, find (f  g)(3)

Solution :

f(x) = 3x - 5 and g(x) = x2

(f  g)(x) = f(g(x))

(f  g)(3) = f(x2)

= f(32)

f(9) = 3(9) – 5

= 27 – 5

= 22

Problem 8 :

If f(x) = -9x - 9 and g(x) = √(x – 9), find (f  g)(10)

Solution :

f(x) = -9x - 9, g(x) = √(x – 9)

(f  g)(10) = ?

 (f ∘ g)(x) = f(g(x))(f ∘ g)(10) = f(g(10))g(10) = √(10–9)= 1 (f ∘ g)(10) = f(1)= -9(1) – 9= -9 – 9= -18(f ∘ g)(10) = -18

Problem 9 :

If f(x) = -4x + 2 and g(x) = √(x – 8), find (f  g)(12)

Solution :

f(x) = -4x + 2, g(x) = √(x – 8)

(f  g)(12) =?

 (f ∘ g)(x) = f(g(x))(f ∘ g)(12) = f(g(12))g(12) = √(12 – 8)g(12) = √4g(12) = 2 (f ∘ g)(12) = f(2)= -4(2) + 2= -8 + 2= -6(f ∘ g)(12) = -6

Problem 10 :

If f(x) = -3x + 4 and g(x) = x2, find (g  f)(-2)

Solution :

f(x) = -3x + 4

g(x) = x2

 (g ∘ f)(x) = g(f(x))(g ∘ f)(-2)= g(f(-2))f(-2) = -3(-2) + 4= 6 + 4= 10 (g ∘ f)(-2)= g(10)= (10)2= 100(g ∘ f)(-2) = 100

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