Inverse of a Quadratic Function and Absolute Value Equations: PECTAA Class 11 Maths, Punjab Board

A parabola fails the horizontal line test, so it has no inverse as it stands. Cut it at the vertex, keep one half, and the inverse appears. The same unit then puts quadratics inside absolute value bars, where every answer has to be checked.

Unit 3 of the PECTAA Mathematics 11 book is Theory of Quadratic Functions. After sketching and maximum or minimum values (section 3.1), section 3.2 finds the inverse of a quadratic function and section 3.3 solves absolute value quadratic equations and inequalities. This note is for FSc Part 1 (1st year, Class 11) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Inverse of a quadratic (section 3.2). A quadratic is usually not one-to-one over its whole domain, so its domain must be restricted to a part where it is one-to-one. Commonly that is x ≥ h or x ≤ h, where h is the x-coordinate of the vertex.

Method.

1. Write y = f(x), then interchange x and y.
2. Solve the new quadratic for y, using the quadratic formula.
3. The result has a + and a − branch. Find the domain and range of f.
4. Keep the branch whose values lie in the range of f. Then domain of f⁻¹ = range of f and range of f⁻¹ = domain of f.

Absolute value (section 3.3). |x| = x when x ≥ 0 and |x| = −x when x < 0.

Absolute value quadratic equations (section 3.3.1). Split |u| = d into u = d or u = −d, solve each, and substitute every answer back into the original equation. Extraneous answers, and here non-real ones, are discarded.

Absolute value quadratic inequalities (section 3.3.2). They take forms such as |ax² + bx + c| < d, > d, ≤ d or ≥ d.

• |u| < d becomes −d < u < d: two inequalities that must both hold (intersection).
• For each quadratic inequality, find the critical values (roots), divide the number line into regions, and test one value in each region.

Key points

• No restriction, no inverse: choose x ≥ h or x ≤ h.
• Domain and range swap between f and f⁻¹.
• Always substitute back in absolute value equations.
• For |u| < d, intersect the two solution sets.

Worked examples

Example 1 (book Example 4). Find the inverse of f(x) = x² + 4x + 3, x ≥ −2, with its domain and range.

1. y = x² + 4x + 3; interchange: x = y² + 4y + 3, so y² + 4y + (3 − x) = 0.
2. y = [−4 ± √(16 − 4(3 − x))]/2 = [−4 ± √(4 + 4x)]/2 = −2 ± √(1 + x).
3. f(x) = (x + 2)² − 1, so for x ≥ −2 the least value is −1. Domain f = [−2, ∞[, range f = [−1, ∞[.
4. Test x = 0: −2 + 1 = −1 lies in [−2, ∞[; −2 − 1 = −3 does not. Discard the minus branch.
5. f⁻¹(x) = −2 + √(1 + x), domain [−1, ∞[, range [−2, ∞[.

Example 2 (book Example 5). Solve |x² − 4| = 5.

1. x² − 4 = 5 gives x² = 9, x = ±3.
2. x² − 4 = −5 gives x² = −1, which has no real solution.
3. Check: |9 − 4| = 5 for both x = 3 and x = −3. Solution set {−3, 3}.

Example 3 (book Example 6). Solve |x² − 6x − 4| < 3.

1. −3 < x² − 6x − 4 < 3.
2. Part (i): x² − 6x − 1 > 0. Roots 3 ± √10, about −0.16 and 6.16. Tests: x = −1 gives 6 > 0 (true); x = 0 gives −1 > 0 (false); x = 7 gives 6 > 0 (true). Solution ]−∞, 3 − √10[ ∪ ]3 + √10, ∞[.
3. Part (ii): x² − 6x − 7 < 0, that is (x + 1)(x − 7) < 0. Tests: x = −2 gives 9 (false); x = 0 gives −7 (true); x = 10 gives 33 (false). Solution ]−1, 7[.
4. Intersection: ]−1, 3 − √10[ ∪ ]3 + √10, 7[.

Common mistakes in the exam

1. Keeping both ± branches of the inverse.
2. Forgetting to swap domain and range for f⁻¹.
3. Not checking absolute value answers.
4. Taking the union instead of the intersection for |u| < d.
5. Rounding critical values too early. Keep 3 ± √10 exact in the final answer.

Exam technique

• For inverses, show the interchange step and the branch test clearly.
• Draw a number line for each quadratic inequality, then a third for the intersection.
• Write final answers in interval notation with exact surds.
• Check your board's notice for the sections in your paper.

Practice questions

1. Find the inverse of f(x) = x² − 3, x ≥ 0. Answer: f⁻¹(x) = √(x + 3), domain [−3, ∞[.
2. Solve |x² − 1| = 3. Answer: x² = 4 gives ±2; x² = −2 has no real root. {−2, 2}.
3. Rewrite |x − 4| < 5 without bars. Answer: −5 < x − 4 < 5, so −1 < x < 9.
4. What is h for f(x) = x² + 6x + 4? Answer: h = −3.
5. Solve |x| = −2. Answer: No solution; an absolute value is never negative.

Ready to practise this? Continue in 1st Year Mathematics.