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Algebra - Functions and Composite Functions

Grade 9IGCSE

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

🔑Concepts

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A function f(x)f(x) is a mapping from a set of inputs (domain) to a set of outputs (range). For every input xx, there is exactly one output yy. This can be visualized using a mapping diagram.

Mapping diagram showing a value x in the domain being mapped to f(x) in the range.
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Composite functions involve applying one function to the result of another. For example, gf(x)gf(x) means you first calculate f(x)f(x) and then use that result as the input for gg.

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The inverse function f−1(x)f^{-1}(x) 'undoes' the operation of f(x)f(x). Graphically, f−1(x)f^{-1}(x) is the reflection of f(x)f(x) in the line y=xy = x.

Graph showing f(x) and its inverse reflected across the line y=x.
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The domain of f(x)f(x) becomes the range of f−1(x)f^{-1}(x), and the range of f(x)f(x) becomes the domain of f−1(x)f^{-1}(x).

📐Formulae

f(x)=yf(x) = y (Basic function notation)

fg(x)=f(g(x))fg(x) = f(g(x)) (Composite function formula)

ff−1(x)=xf f^{-1}(x) = x (Identity property of inverse functions)

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, find f(4)f(4).

Solution:

f(4)=3(4)−5=12−5=7f(4) = 3(4) - 5 = 12 - 5 = 7

Explanation:

Substitute the value 4 into the expression wherever xx appears and simplify.

Problem 2:

If f(x)=2x+1f(x) = 2x + 1 and g(x)=x2g(x) = x^2, find the expression for fg(x)fg(x).

Solution:

fg(x)=f(g(x))=f(x2)=2(x2)+1=2x2+1fg(x) = f(g(x)) = f(x^2) = 2(x^2) + 1 = 2x^2 + 1

Explanation:

To find fg(x)fg(x), substitute the entire expression for g(x)g(x) into the xx position of f(x)f(x).

Problem 3:

Find the inverse function f−1(x)f^{-1}(x) for f(x)=x+32f(x) = \frac{x + 3}{2}.

Solution:

  1. Let y=x+32y = \frac{x + 3}{2}. 2. Swap variables: x=y+32x = \frac{y + 3}{2}. 3. Solve for yy: 2x=y+3⇒y=2x−32x = y + 3 \Rightarrow y = 2x - 3. Therefore, f−1(x)=2x−3f^{-1}(x) = 2x - 3.

Explanation:

The inverse function is found by reversing the operations. We swap xx and yy and isolate yy to find the new rule.

Problem 4:

Given h(x)=5x−2h(x) = 5x - 2, find xx when h(x)=13h(x) = 13.

Solution:

5x−2=13⇒5x=15⇒x=35x - 2 = 13 \Rightarrow 5x = 15 \Rightarrow x = 3

Explanation:

Set the algebraic expression for the function equal to the given value and solve the resulting linear equation for xx.

Problem 5:

Given the functions f(x)=x2f(x) = x^2 and g(x)=x+3g(x) = x + 3, calculate the value of fg(2)fg(2).

Flow diagram showing input 2 being added by 3 to get 5, then squared to get 25.

Solution:

  1. Find g(2)g(2): g(2)=2+3=5g(2) = 2 + 3 = 5
  2. Use the result in f(x)f(x): f(g(2))=f(5)=52=25f(g(2)) = f(5) = 5^2 = 25 Therefore, fg(2)=25fg(2) = 25.

Explanation:

To solve a composite function, start from the inner function and work outwards. Evaluate g(2)g(2) first, then substitute that value into ff.

Problem 6:

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x−43f(x) = \frac{2x - 4}{3} and state its value when x=2x = 2.

Coordinate plot showing the original function and inverse function with specific points highlighted.

Solution:

  1. Let y=2x−43y = \frac{2x - 4}{3}
  2. Swap xx and yy: x=2y−43x = \frac{2y - 4}{3}
  3. Solve for yy: 3x=2y−43x = 2y - 4 3x+4=2y3x + 4 = 2y y=3x+42y = \frac{3x + 4}{2} So, f−1(x)=3x+42f^{-1}(x) = \frac{3x + 4}{2}.
  4. Evaluate for x=2x = 2: f−1(2)=3(2)+42=102=5f^{-1}(2) = \frac{3(2) + 4}{2} = \frac{10}{2} = 5

Explanation:

The inverse function is found by swapping the variables and rearranging the equation to isolate the new yy. This represents the reverse operation.