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Algebra - Composite and inverse functions

Grade 10IB

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

🔑Concepts

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A composite function is formed when the output of one function becomes the input of another. For example, in (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x)), the function gg is applied first, and its result is then passed into function ff. It is helpful to visualize this as a chain of processes.

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The inverse function, denoted as f−1(x)f^{-1}(x), reverses the operation of f(x)f(x). If ff maps xx to yy, then f−1f^{-1} maps yy back to xx. Geometrically, the graph of f−1(x)f^{-1}(x) is a reflection of the graph of f(x)f(x) across the line y=xy = x.

Graph showing f(x), its inverse f^-1(x), and the line of reflection y = x.
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To find the inverse of a function algebraically: 1. Replace f(x)f(x) with yy. 2. Swap xx and yy. 3. Solve the resulting equation for yy. 4. Replace yy with f−1(x)f^{-1}(x).

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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}. This relationship is essential when dealing with restricted domains.

📐Formulae

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

f(f−1(x))=xf(f^{-1}(x)) = x

f−1(f(x))=xf^{-1}(f(x)) = x

If y=f(x), then x=f−1(y)\text{If } y = f(x), \text{ then } x = f^{-1}(y)

💡Examples

Problem 1:

Given f(x)=3x−2f(x) = 3x - 2 and g(x)=x2+1g(x) = x^2 + 1, find the expression for f(g(x))f(g(x)).

Solution:

f(g(x))=3(x2+1)−2f(g(x)) = 3(x^2 + 1) - 2 f(g(x))=3x2+3−2f(g(x)) = 3x^2 + 3 - 2 f(g(x))=3x2+1f(g(x)) = 3x^2 + 1

Explanation:

To find the composite function, we substitute the entire expression of g(x)g(x) into every instance of xx in f(x)f(x).

Problem 2:

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x+53f(x) = \frac{2x + 5}{3}.

Solution:

Step 1: Write as yy: y=2x+53y = \frac{2x + 5}{3} Step 2: Swap xx and yy: x=2y+53x = \frac{2y + 5}{3} Step 3: Solve for yy: 3x=2y+53x = 2y + 5 3x−5=2y3x - 5 = 2y y=3x−52y = \frac{3x - 5}{2} Therefore, f−1(x)=3x−52f^{-1}(x) = \frac{3x - 5}{2}

Explanation:

To find the inverse, we swap the roles of the input and output variables and rearrange the equation to make the new yy the subject.

Problem 3:

If h(x)=5x−4h(x) = 5x - 4, find the value of xx such that h−1(x)=2h^{-1}(x) = 2.

Solution:

If h−1(x)=2h^{-1}(x) = 2, then by the property of inverses: x=h(2)x = h(2) Substitute 22 into the original function: x=5(2)−4x = 5(2) - 4 x=10−4x = 10 - 4 x=6x = 6

Explanation:

Using the property that if f−1(a)=bf^{-1}(a) = b, then f(b)=af(b) = a allows us to solve for xx without finding the inverse expression first.

Problem 4:

Given f(x)=x+3f(x) = x + 3 and g(x)=x2g(x) = x^2, determine the composite function (g∘f)(x)(g \circ f)(x) and calculate the value of (g∘f)(2)(g \circ f)(2). Show the mapping of input x=2x=2.

Mapping diagram showing 2 becoming 5 via f, then 25 via g.

Solution:

  1. Find the expression: (g∘f)(x)=g(f(x))=g(x+3)=(x+3)2(g \circ f)(x) = g(f(x)) = g(x + 3) = (x + 3)^2.
  2. Substitute x=2x = 2: (g∘f)(2)=(2+3)2=52=25(g \circ f)(2) = (2 + 3)^2 = 5^2 = 25.

Explanation:

The inner function ff adds 3 to the input. The outer function gg squares the result. By applying them in sequence, we transform 2 into 5, then 5 into 25.

Problem 5:

Sketch the function f(x)=xf(x) = \sqrt{x} for x≥0x \ge 0 and its inverse f−1(x)=x2f^{-1}(x) = x^2 for x≥0x \ge 0 on the same coordinate plane. Identify the point where they intersect.

Graph of y=sqrt(x) and y=x^2 showing intersection at (1,1).

Solution:

  1. Set f(x)=f−1(x)  ⟹  x=x2f(x) = f^{-1}(x) \implies \sqrt{x} = x^2.
  2. Square both sides: x=x4  ⟹  x4−x=0x = x^4 \implies x^4 - x = 0.
  3. Factor: x(x3−1)=0  ⟹  x=0x(x^3 - 1) = 0 \implies x = 0 or x=1x = 1.
  4. Since x≥0x \ge 0, the intersection points are (0,0)(0,0) and (1,1)(1,1).

Explanation:

Inverse functions are reflections over y=xy=x. Since the square root and squaring functions are inverses (for positive xx), they intersect at points where x=yx = y, specifically at (0,0)(0,0) and (1,1)(1,1).