IB Mathematics AA SL Composite functions Study Notes

IB Mathematics AA SL Composite functions Study Notes

IB Mathematics AA SL Composite functions Study Notes Offer a clear explanation of Composite functions , including various formula, rules, exam style questions as example to explain the topics. Worked Out  examples and common problem types provided here will be sufficient to cover for topic Composite functions

Composite Functions

Composite Functions

A composite function is formed when the output of one function becomes the input of another.

\( (f \circ g)(x) = f(g(x)) \)

Read this  “f composed with g of x”. It means we apply function \( g \) to \( x \), and then apply function \( f \) to the result.

Steps to Evaluate a Composite Function:

  1. Start from the inside: Find \( g(x) \).
  2. Use that result as the input for \( f \), i.e., find \( f(g(x)) \).

Example:

Let the functions be defined as:

  • \( f(x) = 2x + 3 \)
  • \( g(x) = x^2 – 1 \)

Find \( (f \circ g)(x) \) and simplify.

▶️Answer/Explanation

\( g(x) = x^2 – 1 \)

Use that result in \( f(x) \):

\( f(g(x)) = f(x^2 – 1) = 2(x^2 – 1) + 3 \)

\( = 2x^2 – 2 + 3 = 2x^2 + 1 \)

Therefore, \( (f \circ g)(x) = 2x^2 + 1 \)

Identity and Inverse Functions

Identity Function:

The identity function is defined as:

\( I(x) = x \)

 

This function returns the input unchanged. When composing a function with its inverse, the result is the identity function:

\( (f \circ f^{-1})(x) = x \quad \text{and} \quad (f^{-1} \circ f)(x) = x \)

Finding the Inverse Function

To find the inverse function \( f^{-1}(x) \), follow these steps:

  1. Write the function as \( y = f(x) \).
  2. Swap \( x \) and \( y \).
  3. Solve for \( y \). This new equation is \( f^{-1}(x) \).
  4. Check that \( f(f^{-1}(x)) = x \) and \( f^{-1}(f(x)) = x \).

Example: Find the inverse of \( f(x) = 2x + 3 \)

▶️Answer/Explanation

\( y = 2x + 3 \)

Swap \( x \) and \( y \):

\( x = 2y + 3 \)

 Solve for \( y \):

\( x – 3 = 2y \Rightarrow y = \frac{x – 3}{2} \)

So, the inverse is: \( f^{-1}(x) = \frac{x – 3}{2} \)

Verification:

  • \( f(f^{-1}(x)) = 2 \cdot \frac{x – 3}{2} + 3 = x – 3 + 3 = x \)
  • \( f^{-1}(f(x)) = \frac{2x + 3 – 3}{2} = \frac{2x}{2} = x \)

Both compositions return \( x \), confirming that they are inverses.

Inverse Function as a Reflection in the Line \( y = x \)

Geometrically, the graph of the inverse function \( f^{-1}(x) \) is a reflection of the graph of \( f(x) \) across the line \( y = x \).

This means that for every point \( (a, b) \) on the graph of \( f(x) \), there is a corresponding point \( (b, a) \) on the graph of \( f^{-1}(x) \).

The point \( (a, b) \) on \( f \) reflects to \( (b, a) \) on \( f^{-1} \)

Example:

Consider the function \( f(x) = 2x + 1 \).

Its inverse is: \( f^{-1}(x) = \frac{x – 1}{2} \)

The point \( (2, 5) \) lies on the graph of \( f(x) \). Then the point \( (5, 2) \) lies on \( f^{-1}(x) \).

Graphical Interpretation

The reflection across the line \( y = x \) looks like this:

Important Note: Only one-to-one functions have inverses that are also functions. Graphically, this means the function must pass the Horizontal Line Test.

 

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