How to Solve Composite Functions

How to Solve Composite Functions

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Composition of Functions 

Composing functions means combining two or more functions to make a single function.

What is the Composition of Functions?

When \(g(x)\) acts first, the combination of functions \(f(x)\) and \(g(x)\) is shown by \(f(g(x))\) or \((f ∘ g)(x)\). It brings together two or more functions to make a new function. When functions are put together, the output of the function inside the parentheses becomes the input of the function outside the parentheses. So,

For \(f(g(x))\), \(g(x)\) is the input of \(f(x)\)

For \(g(f(x))\), \(f(x)\) is the input of \(g(x)\).

Solving Composition of Functions

When we use BODMAS, we always start by simplifying what is in brackets. So, to find \(f(g(x))\), you have first to figure out \(g(x)\) and plug it into \(f(x)\). In the same way, to find \(g(f(x))\), you must first figure out \(f(x)\) and plug it into \(g(x)\). So, the order is important when finding the composite functions. It means that \(f(g(x))\) might NOT be the same as \(g(f(x))\). With the following steps, we can find the composite function \(f(g(z))\) for any two functions \(f(x)\) and \(g(x)\):

• First, substitute \(x \ = \ z\) in \(g(x)\) to find \(g(z)\).
• Then, substitute \(x \ = \ g(z)\) in \(f(x)\) to find \(f(g(z))\).

With the help of the example below, we can see how these steps work.

Example

Consider: \(f(x) \ = \ 2x^2 \ + \ 5x \ - \ 7\), \(g(x) \ = \ 7x \ - \ 11\)
Find: \(g(f(1))\), \(f(g(2))\)

Solution:

\(g(f(1)) \ = \ g(2(1)^2 \ + \ 5(1) \ - \ 7) \ = \ g(0) \ = \ 7(0) \ - \ 11 \ = \ -11\)

\(f(g(2)) \ = \ f(7(2) \ - \ 11) \ = \ f(3) \ = \ 2(3)^2 \ + \ 5(3) \ - \ 7 \ = \ 26\)

The Domain of Composite Functions

If \(g: \ x \ → \ y\) and \(f: \ y \ → \ z\), then \(f ∘ g: \ x \ → \ z\). In other words, \(f ∘ g\)'s domain is \(x\), and its range is \(z\). Here are the steps to find the domain of the composite function \(f ∘ g\) when the functions are defined algebraically.

• Find the domain of \(g(x)\), the inner function (Let this be \(A\))
• Calculate \(f(g(x))\) to finding the domain of the function (let's say it's \(B\)).
• Find the intersection of \(A\) and \(B\), and \(A \ ∩ \ B\) will give you the domain of \(f(g(x))\).

Example:

Consider \(f(x) \ = \ \frac{5}{2x^2 \ + \ 3}\) and \(g(x) \ = \ \sqrt{2x \ - \ 4}\), find the domain of \(f(g(x))\).

Solution:

The inner function is \(g(x)\), and its domain is:

\(2x \ - \ 4 \ ≥ \ 0 \ ⇒ \ 2x \ ≥ \ 4 \ ⇒ \ x \ ≥ \ 2\) (The expression under the radical cannot be negative.)

So, \(A \ = \ \{x \ | \ x \ ≥ \ 2 \}\) OR \(A \ = \ [2 \ , \ ∞)\).

Now we'll figure out \(f(g(x))\).

\(f(g(x)) \ = \ f(\sqrt{2x \ - \ 4}) \ = \ \frac{5}{2(2x \ - \ 4)^2 \ + \ 3} \ = \ \frac{5}{2(\sqrt{2x \ - \ 4})^2 \ + \ 3} \ = \ \frac{5}{4x \ - \ 8 \ + \ 3} \ = \ \frac{5}{4x \ - \ 5}\)

So, the domain of \(f(g(x))\) is: \(A \ = \ \{x \ | \ x \ ≠ \ \frac{5}{4} \} \ \) (The denominator cannot be zero).

So, the domain of \(f(g(x))\) is, \(A \ ∩ \ B \ = \ \{x \ | \ x \ ≥ \ 2 \}\) OR \(A \ = \ [2 \ , \ ∞)\).

The Range of Composite Functions

The range of a composite function is found the same way that the range of any other function is found. It doesn't matter if the functions are inside or outside. Let's figure out the \(f(g(x))\)'s range in the last example. We got \(f(g(x)) \ = \ \frac{5}{4x \ - \ 5}\) or we can say \(y \ = \ \frac{5}{4x \ - \ 5}\). So, to find the range, we solve for \(x\) and ensure the denominator is not equal to zero.

\(y \ = \ \frac{5}{4x \ - \ 5} \ ⇒ \ (4x \ - \ 5)y \ = \ 5 \ ⇒ \ 4xy \ - \ 5y \ = \ 5 \ ⇒ \ 4xy \ = \ 5 \ + \ 5y \ ⇒ \ x \ = \ \frac{5 \ + \ 5y}{4y}\)

\(4y \ ≠ \ 0\) gives \(y \ ≠ \ 0\) for the range. So, the range is: \(\{y \ | \ y \ ≠ \ 0\}\).

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