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Algebra and Graphs - Function notation and inverse functions

Grade 11IGCSE

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

🔑Concepts

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Function Notation: f(x)f(x) represents the value of the function ff at the input xx. It is read as 'f of x'.

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Domain and Range: The domain is the set of all possible input values (xx), and the range is the set of all possible output values (yy).

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Composite Functions: fg(x)fg(x) means applying function gg first, then applying function ff to the result: f(g(x))f(g(x)). Order matters; fg(x)≠gf(x)fg(x) \neq gf(x) usually.

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Inverse Functions: f−1(x)f^{-1}(x) reverses the action of f(x)f(x). If f(a)=bf(a) = b, then f−1(b)=af^{-1}(b) = a.

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

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Existence of Inverse: A function has an inverse only if it is a one-to-one mapping.

📐Formulae

f(x)=yf(x) = y

fg(x)=f(g(x))fg(x) = f(g(x))

f2(x)=f(f(x))f^2(x) = f(f(x))

f(f−1(x))=xf(f^{-1}(x)) = x

To find f−1(x)f^{-1}(x): 1. Let y=f(x)y = f(x), 2. Swap xx and yy, 3. Rearrange to make yy the subject.

💡Examples

Problem 1:

Given f(x)=3x−5f(x) = 3x - 5 and g(x)=x2+1g(x) = x^2 + 1, find fg(2)fg(2).

Solution:

  1. Find g(2)=(2)2+1=4+1=5g(2) = (2)^2 + 1 = 4 + 1 = 5.
  2. Find f(g(2))=f(5)=3(5)−5=15−5=10f(g(2)) = f(5) = 3(5) - 5 = 15 - 5 = 10.

Explanation:

To solve a composite function, evaluate the inner function first, then substitute that result into the outer function.

Problem 2:

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x+3x−1f(x) = \frac{2x + 3}{x - 1}.

Solution:

  1. Let y=2x+3x−1y = \frac{2x + 3}{x - 1}.
  2. Swap xx and yy: x=2y+3y−1x = \frac{2y + 3}{y - 1}.
  3. Multiply both sides: x(y−1)=2y+3x(y - 1) = 2y + 3.
  4. Expand: xy−x=2y+3xy - x = 2y + 3.
  5. Group yy terms: xy−2y=x+3xy - 2y = x + 3.
  6. Factor out yy: y(x−2)=x+3y(x - 2) = x + 3.
  7. Solve for yy: y=x+3x−2y = \frac{x + 3}{x - 2}.
  8. Therefore, f−1(x)=x+3x−2f^{-1}(x) = \frac{x + 3}{x - 2}.

Explanation:

To find an inverse, we treat f(x)f(x) as yy, swap variables to reverse the relationship, and use algebraic manipulation to isolate the new yy.

Problem 3:

If h(x)=5x−2h(x) = 5x - 2, solve the equation h(x)=h−1(8)h(x) = h^{-1}(8).

Solution:

  1. Find h−1(8)h^{-1}(8) by setting h(x)=8h(x) = 8: 5x−2=8  ⟹  5x=10  ⟹  x=25x - 2 = 8 \implies 5x = 10 \implies x = 2. So, h−1(8)=2h^{-1}(8) = 2.
  2. Set h(x)=2h(x) = 2: 5x−2=25x - 2 = 2.
  3. Solve for xx: 5x=4  ⟹  x=0.85x = 4 \implies x = 0.8.

Explanation:

Instead of finding the full expression for h−1(x)h^{-1}(x), you can use the property that if h(a)=bh(a) = b, then h−1(b)=ah^{-1}(b) = a to find numerical values quickly.