krit.club logo

Algebra and Graphs - Function notation and inverse functions

Grade 11A Level

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

🔑Concepts

•

A function f(x)f(x) maps an input xx to a unique output yy. Mapping diagrams visualize this relationship by showing how elements from the domain (input) relate to elements in the codomain (output). For f(x)=x+2f(x) = x + 2, each input is shifted by 22.

Mapping diagram showing values being mapped from a domain to a range by adding 2.
•

The inverse function f−1(x)f^{-1}(x) reverses the operation of f(x)f(x). Graphically, 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. If a point (a,b)(a, b) lies on ff, then (b,a)(b, a) lies on f−1f^{-1}.

Graph showing f(x), its inverse, and the line of reflection y=x.
•

Composite functions like fg(x)fg(x) mean applying gg first, and then applying ff to the result. f2(x)f^2(x) is shorthand for f(f(x))f(f(x)). The order is crucial: fg(x)fg(x) is generally not equal to gf(x)gf(x).

•

To find the inverse algebraically: 1. Replace f(x)f(x) with yy. 2. Swap xx and yy. 3. Solve for the new yy. The resulting expression is f−1(x)f^{-1}(x).

📐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.

Problem 4:

Given the function f(x)=12x+1f(x) = \frac{1}{2}x + 1, find f−1(x)f^{-1}(x) and sketch both functions on the same axes to show the reflection in y=xy = x.

Graph of y = 0.5x + 1 and y = 2x - 2.

Solution:

y=12x+1y = \frac{1}{2}x + 1 Swap xx and yy: x=12y+1x = \frac{1}{2}y + 1 Subtract 1: x−1=12yx - 1 = \frac{1}{2}y Multiply by 2: y=2(x−1)y = 2(x - 1) f−1(x)=2x−2f^{-1}(x) = 2x - 2

Explanation:

To find the inverse, we isolate the variable that was originally the input. The graph confirms that f(x)f(x) and f−1(x)f^{-1}(x) are mirror images across the line y=xy = x.

Problem 5:

The function g(x)=x2g(x) = x^2 is defined for x≥0x \ge 0. Find the value of g−1(9)g^{-1}(9) and illustrate the mapping.

Flow diagram showing 3 mapping to 9 under g and 9 mapping back to 3 under g inverse.

Solution:

To find g−1(9)g^{-1}(9), we set g(x)=9g(x) = 9: x2=9x^2 = 9 Since x≥0x \ge 0: x=9=3x = \sqrt{9} = 3 So, g−1(9)=3g^{-1}(9) = 3.

Explanation:

An inverse function maps the output back to the input. For g(x)=x2g(x)=x^2, the inverse operation for positive values is the square root.