EVALUATING THE VALUE OF COMPOSITION FUNCTION FROM THE GIVEN FUNCTIONS

Problem 1 :

Evaluate the indicated expression assuming that f(x) = √x, g(x) = (x + 1)/(x + 2) , h(x) = |x - 1|.

(i)  (f o g) (4)

(ii)  (f o g) (5)

(iii)  (g o f) (4)

(iv)  (g o f) (5)

(v)  (f o g o h) (0)

(vi)  (h o g o f) (0)

Solution :

(i)  (f o g) (4) :

(f o g) (4) = f[g(4)] ----(1)

g(x) = (x + 1)/(x + 2)

g(4) = (4 + 1)/(4 + 2) = 5/6

Substitute g(4) = 5/6 in (1).

(f o g) (4) = f(5/6)

= √(5/6)

(ii)  (f o g) (5) :

(f o g) (5) = f[g(5)] ----(2)

g(x) = (x + 1)/(x + 2) 

g(5) = (5 + 1)/(5 + 2) = 6/7

Substitute g(5) = 6/7 in (2).

(f o g) (5) = f(6/7)

= √x

= √(6/7)

(iii)  (g o f)(4) : 

(g o f) (4) = g[f(4)] ----(3)

f(x) = √x

f(4) = √4 = 2

Substitute f(4) = 2 in (3).

(g o f) (4) = g(2)

= (2 + 1)/(2 + 2)

= 3/4

(iv)  (g o f) (5) :

(g o f) (5) = g[f(5)] ----(4)

f(x) = √x

f(5) = √5

Substitute f(5) = √5 in (4)

(g o f) (5) = g[√5]

= (√5 + 1)/(√5 + 2)

(v)  (f o g o h) (0) :

(f o g o h) (0) = (f o g) [h(0)] ----(5)

h(x) = |x - 1|

h(0) = |0 - 1| = 1

Substitute h(0) = 1 in (5). 

(f o g o h) (0) = (f o g) (1)

= f[g(1)] ----(6)

g(x) = (x + 1)/(x + 2)

g(1) = (1 + 1)/(1 + 2)  =  2/3

Substitute g(1) = 2/3 in (6). 

= f[2/3]

= √(2/3)

(vi)  (h o g o f) (0) :

(h o g o f) (0) = (h o g) [f(0)] ----(7)

f(x) = √x

f(0) = √0 = 0

Substitute f(0) = 0 in (7). 

  = (h o g) (0)

= h[g(0)] ----(8)

g(x) = (x + 1)/(x + 2)

g(0) = (0 + 1)/(0 + 2)  =  1/2

Substitute g(0) = 1/2 in (8). 

  = h[1/2]

= |1/2 − 1|

= |(1 - 2)/2|

= |-1/2|

= 1/2

Problem 2 :

let f (x) = 3x + 5, g(x) = x², and h(x) = −2x − 1.

Perform the indicated composition.

a. f (g(h(x)))       b. h(g(f (x)))

c. f (f (f (x)))      d. g(h(g(x)))

Solution :

f (x) = 3x + 5, g(x) = x², and h(x) = −2x − 1.

a. f (g(h(x))) :

g(h(x)) = g(-2x - 1)

= (-2x - 1)²

= 4x² - 2(-2x)(1) + 1²

= 4x² + 4x + 1

f (g(h(x))) = 3(4x² + 4x + 1) + 5

= 12x² + 12x + 3 + 5

= 12x² + 12x + 8

b. h(g(f (x)))

g(f(x)) = g(3x + 5)

= (3x + 5)²

= (3x)² + 2(3x)(5) + 5²

= 9 x² + 30x + 25

h(g(f (x))) = -2(9 x² + 30x + 25) - 1

= -18x² - 60x - 50 - 1

= -18x² - 60x - 51

c. f (f (f (x))) :

f(f(x)) = f(3x + 5)

= 3(3x + 5)

= 9x + 15

f(f(f(x))) = 3(9x + 15)

= 27x + 45

d. g(h(g(x))) :

h(g(x)) = h(x²)

h(g(x)) = −2x² − 1.

g(h(g(x))) = g(−2x² − 1)

=  (−2x² − 1)²

= (-2x²)² - 2(−2x²)(1) + 1²

= 4x⁴ + 4x² + 1

Problem 3 :

The function p(d) = 0.03d + 1 approximates the pressure (in atmospheres) at a depth of d feet below sea level. The function d(t) = 60t represents the depth (in feet) of a diver t minutes after beginning a descent from sea level, where 0 ≤ t ≤ 2.

a. Find p(d(t)). Interpret the terms and coefficient.

b. Evaluate p(d(1.5)) and explain what it represent

Solution :

a)

p(d) = 0.03d + 1

d(t) = 60t

p(d(t)) = p(60t)

p(d(t)) = 0.03(60t) + 1

= 1.8t + 1

b) p(d(1.5))

d(1.5) = 60(1.5)

= 4

P(4) =  0.03(4) + 1

= 0.12 + 1

= 1.12

So, the value of  is 1.12.

The number of bacteria in a refrigerated food product is given by

N(T) =  23T² - 56 T + 1

 where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by 

T(t) = 5t + 1.5

where t is the time in hours.

a. Find the composite function N(T(t)

b. Find the bacteria count after 4 hours

Solution :

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