EVALUATING A COMPOSITE FUNCTION

Example 1 :

Let f(x) = 2x² - 3 and g(x) = 4x.

Evaluate the following :

(i) f ∘ g(1)

(ii) g ∘ f(1)

(iii) (f ∘ f)(-2)

(iv) (g ∘ g)(-1)

Solution :

(i) f ∘ g(1) :

f ∘ g(1) = f[g(1)]

= f[4(1)]

= f(4)

= 2(4)² - 3

= 2(16) - 3

= 32 - 3

= 29

(ii) g ∘ f(1) :

g ∘ f(1) = g[2(1)² - 3]

= g[2(1) - 3]

= g(2 - 3)

= g(-1)

= 4(-1)

= -4

(iii) f ∘ f(-2) :

f ∘ f(-2) = f[2(-2)² - 3]

= f[2(4) - 3]

= f(8 - 3)

= f(5)

= 2(5)² - 3

= 2(25) - 3

= 50 - 3

= 47

(iii) g ∘ g(-1) :

g ∘ g(-1) = g[4(-1)]

= g(-4)

= 4(-4)

= -16

Example 2 :

Let f(x) = 2x - 1, g(x) = 3x and h(x) = x² + 1.

Evaluate the following :

(i) f ∘ g(-3)

(ii) f ∘ h(7)

(iii) (g ∘ h)(24)

(iv) f{g[h(2)]}

Solution :

f(x) = 2x - 1, g(x) = 3x and h(x) = x² + 1

(i) f ∘ g(-3) :

f ∘ g(-3) = f[g(-3)]

= f[3(-3)]

= f(-9)

= 2(-9) - 1

= -18 - 1

= -19

(ii) f ∘ h(7) :

f ∘ h(7) = f[h(7)]

= f(7² + 1)

= f(49 + 1)

= f(50)

= 2(50) - 1

= 100 - 1

= 99

(iii) g ∘ h(24) :

g ∘ h(24) = g[h(24)]

= g(24² + 1)

= g(576 + 1)

= g(577)

= 3(577)

= 1731

(iv) f{g[h(2)]} :

f{g[h(2)]} = f{g[2² + 1]}

= f{g[4 + 1]}

= f{g(5)}

= f{3(5)}

= f(15)

= 2(15) - 1

= 30 - 1

= 29

Example 3 :

Let f(x) = -4x + 2 and g(x) = √(x - 8).

Evaluate : f ∘ g(12).

Solution :

f(x) = -4x + 2 and g(x) = √(x - 8)

f ∘ g(12) = f[g(12)]

= f[√(12 - 8)]

= f(√4)

= f(2)

= -4(2) + 2

= -8 + 2

= -6

Example 4 :

Let f(x) = -2x + 1 and g(x) = √(x² - 5).

Evaluate : g ∘ f(2).

Solution :

f(x) = -2x + 1 and g(x) = √(x² - 5)

g ∘ f(2) = g[f(2)]

= g[-2(2) + 1]

= g(-4 + 1)

= g(-3)

= √[(-3)² - 5]

= √(9 - 5)

= √4

= 2

Example 5 :

Let f(x) = log₁₀x and g(x) = 10^x.

Evaluate : f ∘ g(5).

Solution :

f(x) = log₁₀x and g(x) = 10^x

f ∘ g(5) = f[g(5)]

= f(10⁵)

= log₁₀(10⁵)

= 5log₁₀(10)

= 5(1)

= 5

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