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Algebra - Functions (Composite and Inverse)

Grade 12A Level

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

🔑Concepts

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A function f:A→Bf: A \to B maps an input xx from the domain to an output yy in the range. The inverse function f−1f^{-1} reverses this process, mapping yy back to xx. This is only possible if the function is one-to-one (bijective).

Flowchart showing the relationship between a function and its inverse.
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The graph of an inverse function f−1(x)f^{-1}(x) is a reflection of the graph of f(x)f(x) in the line y=xy = x. If a point (a,b)(a, b) lies on f(x)f(x), then the point (b,a)(b, a) lies on f−1(x)f^{-1}(x).

Graph showing f(x) and its inverse reflected across the line y=x.
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Composite functions involve applying one function to the result of another. For fg(x)fg(x), the inner function g(x)g(x) is evaluated first, and its output becomes the input for the outer function ff. Note that fg(x)≠gf(x)fg(x) \neq gf(x) in most cases.

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To solve equations involving composites, such as fg(x)=kfg(x) = k, first define the expression for fg(x)fg(x) by substituting g(x)g(x) into f(x)f(x), then solve the resulting algebraic equation for xx.

Visual mapping of nested function domains.

📐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.

Problem 4:

Given the functions f(x)=x2−4f(x) = x^2 - 4 and g(x)=x+4g(x) = \sqrt{x + 4} for x≥−4x \geq -4, prove that g(x)g(x) is the inverse of f(x)f(x) for the domain x≥0x \geq 0 by finding fg(x)fg(x) and gf(x)gf(x).

Graphs of f(x) and g(x) showing symmetry over y=x.

Solution:

  1. Find fg(x)fg(x): fg(x)=f(g(x))=(x+4)2−4fg(x) = f(g(x)) = (\sqrt{x + 4})^2 - 4 fg(x)=x+4−4=xfg(x) = x + 4 - 4 = x
  2. Find gf(x)gf(x): gf(x)=g(f(x))=(x2−4)+4gf(x) = g(f(x)) = \sqrt{(x^2 - 4) + 4} gf(x)=x2=xgf(x) = \sqrt{x^2} = x (since x≥0x \geq 0) Since fg(x)=gf(x)=xfg(x) = gf(x) = x, g(x)=f−1(x)g(x) = f^{-1}(x).

Explanation:

If the composition of two functions results in the identity function xx, the functions are inverses of each other. The domain restriction x≥0x \geq 0 ensures f(x)f(x) is one-to-one.

Problem 5:

Let f(x)=3x−2f(x) = 3x - 2 and h(x)=2x+kh(x) = 2x + k. Find the value of the constant kk such that fh(x)=hf(x)fh(x) = hf(x).

Comparison of two different composition paths to find k.

Solution:

  1. Calculate fh(x)fh(x): fh(x)=f(2x+k)=3(2x+k)−2fh(x) = f(2x + k) = 3(2x + k) - 2 fh(x)=6x+3k−2fh(x) = 6x + 3k - 2
  2. Calculate hf(x)hf(x): hf(x)=h(3x−2)=2(3x−2)+khf(x) = h(3x - 2) = 2(3x - 2) + k hf(x)=6x−4+khf(x) = 6x - 4 + k
  3. Set them equal: 6x+3k−2=6x−4+k6x + 3k - 2 = 6x - 4 + k 3k−2=−4+k3k - 2 = -4 + k 2k=−22k = -2 k=−1k = -1

Explanation:

To make the composition commutative, we find expressions for both orders of composition and equate the constant terms since the coefficients of xx are already equal.