# EVALUATING COMPOSITOION OF FUNCTION FROM TABLE

If  f and g are functions, then the composite function of f and g is defined by

(f ∘ g)(x) = f( g(x) )

The domain of (f ∘ g) is the set of all real numbers x in the domain of g such that ݃g(x) is in the domain of f.

Problem 1 :

Use the table of values to evaluate each expression

 1)  f ( g(8) )4)  g ( f (3) )7)  g ( g (2) ) 2)  f( g(5) )5)  f ( f (4) )8)  g ( g (6) ) 3)  g ( f (5) )6)  f ( f (1) )9)  g ( g (1) )

Solution :

1)  f ( g(8) )

From the table, value of g(8) is 4.

f ( g(8) ) = f(4)

= 4

2)  f ( g(5) )

From the table, value of g(5) is 8.

f ( g(5) ) = f(8)

= 9

3)  g ( f (5) )

From the table, value of f(5) is 0.

g ( f(5) ) = f(0)

= 7

4)  g ( f (3) )

From the table, value of f(3) is 8.

g ( f(3) ) = f(8)

= 9

5)  f ( f (4) )

From the table, value of f(4) is 4.

f ( f(4) ) = f(4)

= 4

6) f ( f (1) )

From the table, value of f(1) is 6.

f ( f(1) ) = f(6)

= 2

7) g ( g (2) )

From the table, value of g(2) is 6.

f ( f(1) ) = f(6)

= 2

8) g ( g (6) )

From the table, value of g(6) is 7.

g ( g(6) ) = g(7)

= 1

9)  g ( g (1) )

From the table, value of g(1) is 5.

g ( g(1) ) = g(5)

= 8

Problem 2 :

Evaluating Composite Functions :

The tables give some selected ordered pairs for functions f and g

 1) (f ∘ g)(2)2) (f ∘ g)(1)3) (f ∘ g)(9) 4) (f ∘ g)(5)5) (g ∘ f)(6)6) (f ∘ g)(7) 7) (g ∘ f)(12)8) (g ∘ f)(3)

Solution :

1) (f ∘ g)(2)

(f ∘ g)(2) = f [ g(2) ] ----(1)

The value of g(2) is 3.

Applying it in (1), we get

= f [ 3 ]

(f ∘ g)(2) = 9

2) (f ∘ g)(1)

(f ∘ g)(1) = f [ g(1) ] ----(1)

The value of g(1) is 9.

Applying it in (1), we get

= f [ 9 ]

(f ∘ g)(9) = 6

3) (f ∘ g)(9)

(f ∘ g)(9) = f [ g(9) ] ----(1)

The value of g(9) is 12.

Applying it in (1), we get

= f [ 12 ]

(f ∘ g)(12) = 5

4) (f ∘ g)(5)

(f ∘ g)(5) = f [ g(5) ] ----(1)

The value of g(5) is 6.

Applying it in (1), we get

= f [ 6 ]

(f ∘ g)(6) = 7

5) (g ∘ f)(6)

(g ∘ f)(6) = g [ f(6) ] ----(1)

The value of f(6) is 7.

Applying it in (1), we get

= f [ 7 ]

(g ∘ f)(6) = -3

6) (f ∘ g)(7)

(f ∘ g)(7) = f [ g(7) ] ----(1)

The value of g(7) is 4.

Applying it in (1), we get

= f [ 4 ]

(f ∘ g)(7) = -3

7) (g ∘ f)(12)

(g ∘ f)(12) = g [ f(12) ] ----(1)

The value of f(12) is 5.

Applying it in (1), we get

= g [ 5 ]

(g ∘ f)(12) = 6

8) (g ∘ f)(3)

(g ∘ f)(3) = g [ f(3) ] ----(1)

The value of f(3) is 9.

Applying it in (1), we get

= g [ 9 ]

(g ∘ f)(3) = 12

Problem 3 :

Use the tables of ordered pairs to determine the value of each composite function

 1) (𝑓 ∘ 𝑔)(36) =2)  (𝑔 ∘ 𝑔)(16) = 3) (𝑔 ∘ 𝑓)(4) =4)  (𝑓 ∘ 𝑓)(4) =

Solution :

 1) (𝑓 ∘ 𝑔)(36)= 𝑓 [ 𝑔(36) ]= f[6]= 21 2)  (𝑔 ∘ 𝑔)(16) == g [ 𝑔(16) ]= g[4]= 2 3) (𝑔 ∘ 𝑓)(4) == g [ f(4) ]= g[1]= 1 4) (𝑓 ∘ 𝑓)(4) == f [ f(4) ]= f[1]= -14

Problem 4 :

The table below shows values of 𝑓(𝑥) at selected values of 𝑥. The function 𝑔(𝑥) is shown in the graph below.

Let h be the function defined by ℎ(𝑥) = 2|𝑥 − 4|.

Find:

 1. 𝑦 = ℎ(𝑓(2))2. 𝑦 = ℎ(𝑔(3))3. 𝑦 = 𝑔(𝑓(−2)) 4. 𝑦 = 𝑓(𝑔(−3))5. 𝑦 = 𝑔(𝑓(ℎ(3)))6. Find ℎ(𝑓(𝑔(0)))

Solution :

1. 𝑦 = ℎ(𝑓(2))

From the table value of f(2) is 5.

y = h(5)

Given ℎ(𝑥) = 2|𝑥 − 4|

h(5) = 2|5 - 4|

y = h(5) = 2

2. 𝑦 = ℎ(𝑔(3))

From the graph, the value of g(3) is -1.

y = h(-1)

Given ℎ(𝑥) = 2|𝑥 − 4|

h(-1) = 2|-1 - 4|

y = h(-1) = 10

3. 𝑦 = 𝑔(𝑓(−2))

y = g(5)

From the table, value of f(-2) is 5.

y = g(5)

y = 3

4. 𝑦 = 𝑓(𝑔(−3))

From the graph, value of g(-3) is 2.

𝑦 = 𝑓(2)

From the table, the value of f(2) is 5.

y = 5

5. 𝑦 = 𝑔(𝑓(ℎ(3)))

From the given function, value of h(3) is

ℎ(3) = 2|3 − 4|

h(3) = 2

𝑦 = 𝑔(𝑓(2))

Value of f(2) is 5.

𝑦 = 𝑔(5)

From the table, the value of g(5) is 3.

y = 3

6. Find ℎ(𝑓(𝑔(0)))

From the graph, the value of g(0) = 0

y = ℎ(𝑓(0))

From the table, the value of f(0) is 1.

𝑦 = h(1)

ℎ(1) = 2|1 − 4|

h(2) = 6

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