COMPOSITION OF A SQUARE ROOT FUNCTION

Example 1 :

Let f(x) = x² - 1 and let g(x) = √x. Find the domain of the composition functions

(i)  g∘f 

(ii)  f∘g 

Solution :

Domain of f(x) :

f(x) = x² - 1

All real values, so domain is (-∞, ∞)

Domain of g(x) :

g(x) = √x

All positive values, so domain is [0, ∞)

(i)  g∘f 

g∘f  = g( f(x) )

  = g(x² - 1)

Here instead of x, we have  x² - 1. So, in the function g(x) we have to apply x² - 1.

g∘f  = √(x² - 1).

Domain of g∘f :

√(x² - 1)  ≥ 0

(x² - 1)  ≥ 0

(x + 1) (x - 1)  ≥ 0

x  ≥ 1 and  x ≥ -1

Comparing domain of f(x) and this domain the intersection part is (-∞, -1] and [1, ∞).

So the required domain for  g∘f is (-∞, -1] U [1, ∞).

(ii)  f∘g 

f∘g  = f( g(x) )

  = f(√x)

Here we see √x instead of x, so we will apply x as √x in the function f(x).

f∘g = (√x)² - 1

= x - 1

Domain of f∘g :

Domain of x - 1 is all real values.

Comparing domain of g(x) and this domain the intersection part is [0, ∞).

So, the domain f∘g is [0, ∞). 

Example 2 :

Let

f(x) = 1/(x² - 1) and g(x) = √(x - 2)

Find the domain of f(g(x)).

Solution :

Domain of g(x) is real values ≥ 2

Domain of 1/(x - 3) is all real values except 3.

So, the required domain is f(g(x)) is [2, 3) U (3, ∞).

Example 3 :

Let

f(x) = √(x + 2) and g(x) = x²

Find the domain of f(g(x)).

Solution :

f(g(x)) = f(x²)

Here instead of x, we have x². So in the function f(x), we have to apply the value x²

= √(x² + 2)

Even though we give negative values of x, f(g(x)) will give us positive result.

So, the domain of f(g(x)) is all real values.

Example 4 :

The function D(p) gives the number of items that will be demanded when the price is p. The product cost C(x), is the cost of producing x items. To determine the cost of production when the price is $6, you would do which of the following.

a)  Evaluate D(C(6))      b)  Evaluate C(D(6))

c)  Solve D(C(x)) = 6      d)  Solve C(D(p)) = 6

Solution :

Price of item = p

Number of items = x

Cost of production when price = 6

p = 6

Number of items demanded = D(p)

Product cost of x items = C(x)

Evaluating C(D(6)), we get to know the cost of production when the price is $6.

Example 5 :

The function A(d) gives the pain level of a scale of 0-10 experienced by a patient with d milligrams of a pain reduction drug in their system. The milligrams of drug in the patient's system after t minutes is modeled by m(t). To determine when the patient will be a pain level 4, you would need to 

a)  Evaluate A(m(4))      b)  Evaluate m(A(4))

c)  Solve A(m(t)) = 4      d)  Solve m(A(d)) = 4

Solution :

After intake the miilgram of drug into the patient's system, then he can experience level 4 pain.

Example 6 :

The radius r in inches of a spherical ballon is related to the Volume V, by 

r(V) = ∛(3V/4 π)

Air is pumped into a balloon, so the volume after t seconds is given by V(t) = 10 + 20t

a) Find the composite function r(V(t))

b)  Find the radius after 20 seconds.

Solution :

a) 

r(V) = ∛(3V/4 π)

V(t) = 10 + 20t

r(V(t)) = ∛(3(10+20t) / 4 π)

b)

Applying t = 20

r(V(20)) = ∛(3(10+20(20)) / 4 π)

= ∛(3(410) / 4 π)

= ∛97.92

= 4.69 inches

Example 7 :

A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to

r(t) = 25√(t + 2)

find the area of the ripple as a function of time. Find the area of the ripple at t = 2.

Solution :

r(t) = 25√(t + 2)

When t = 2

r(2) = 25√(2 + 2)

r(2) = 25√4

= 25(2)

= 50

So, area of the ripple is 50 square inches

Example 8 :

For the following exercises, find the composition when

f(x) = x² + 2 for all x ≥ 0 and g(x) = √(x − 2)

a) (f ∘ g) (6), (g ∘ f) (6)

b) (g ∘ f) (a), (f ∘ g) (a)

c) (f ∘ g) (11), (g ∘ f) (11)

f(x) = x² + 2 for all x ≥ 0 and g(x) = √(x − 2)

a) (f ∘ g) (6) = 6 and (g ∘ f) (6) = 6

b) (g ∘ f) (a) = a, (f ∘ g) (a) = a

c) (f ∘ g) (11) = 11, (g ∘ f) (11) = 11