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Algebra - Functions (Composite and Inverse)

Grade 12IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of a Function: A mapping where every input (domain) has exactly one output (range).

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Composite Functions: The application of one function to the result of another, denoted as fg(x)=f(g(x))fg(x) = f(g(x)).

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Order of Operations: In fg(x)fg(x), the function gg is applied first, followed by ff.

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Inverse Function (f−1f^{-1}): A function that reverses the effect of ff. It exists only if the function is one-to-one.

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Domain and Range: The domain of ff becomes the range of f−1f^{-1}, and the range of ff becomes the domain of f−1f^{-1}.

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Graphical Relationship: The graph of y=f−1(x)y = f^{-1}(x) is a reflection of y=f(x)y = f(x) in the line y=xy = x.

📐Formulae

fg(x)=f(g(x))fg(x) = f(g(x)) (Composite Function)

f(f−1(x))=xf(f^{-1}(x)) = x (Identity Property)

f−1(f(x))=xf^{-1}(f(x)) = x (Identity Property)

To find f−1(x)f^{-1}(x): Set y=f(x)y = f(x), swap xx and yy, then solve for yy.

💡Examples

Problem 1:

Given f(x)=2x+3f(x) = 2x + 3 and g(x)=x2g(x) = x^2, find fg(x)fg(x) and gf(x)gf(x).

Solution:

fg(x)=f(g(x))=f(x2)=2(x2)+3=2x2+3fg(x) = f(g(x)) = f(x^2) = 2(x^2) + 3 = 2x^2 + 3. gf(x)=g(f(x))=g(2x+3)=(2x+3)2=4x2+12x+9gf(x) = g(f(x)) = g(2x+3) = (2x+3)^2 = 4x^2 + 12x + 9.

Explanation:

Substitute the entire expression of the inner function into every instance of 'x' in the outer function. Note that fg(x)≠gf(x)fg(x) \neq gf(x) in most cases.

Problem 2:

Find the inverse of the function f(x)=3x−1x+2f(x) = \frac{3x - 1}{x + 2} where x≠−2x \neq -2.

Solution:

  1. Let y=3x−1x+2y = \frac{3x - 1}{x + 2}
  2. Swap xx and yy: x=3y−1y+2x = \frac{3y - 1}{y + 2}
  3. Multiply by (y+2)(y+2): x(y+2)=3y−1x(y + 2) = 3y - 1
  4. Expand: xy+2x=3y−1xy + 2x = 3y - 1
  5. Rearrange to group yy: xy−3y=−2x−1xy - 3y = -2x - 1
  6. Factor out yy: y(x−3)=−(2x+1)y(x - 3) = -(2x + 1)
  7. Solve for yy: y=−(2x+1)x−3=2x+13−xy = \frac{-(2x + 1)}{x - 3} = \frac{2x + 1}{3 - x}. Therefore, f−1(x)=2x+13−xf^{-1}(x) = \frac{2x + 1}{3 - x}.

Explanation:

The method involves switching the roles of xx and yy and using algebraic manipulation to isolate the new yy as the subject.

Problem 3:

If f(x)=5x−2f(x) = 5x - 2, solve the equation f−1(x)=f(1)f^{-1}(x) = f(1).

Solution:

  1. Find f(1)f(1): f(1)=5(1)−2=3f(1) = 5(1) - 2 = 3.
  2. Find f−1(x)f^{-1}(x): y=5x−2⇒x=5y−2⇒y=x+25y = 5x - 2 \Rightarrow x = 5y - 2 \Rightarrow y = \frac{x+2}{5}. So f−1(x)=x+25f^{-1}(x) = \frac{x+2}{5}.
  3. Set them equal: x+25=3\frac{x+2}{5} = 3.
  4. Solve: x+2=15⇒x=13x + 2 = 15 \Rightarrow x = 13.

Explanation:

Calculate the numerical value of f(1)f(1) first, then equate it to the derived inverse function to solve for the unknown xx.