Learning Objectives
Connect evaluation, domain, range, composition, and inversion.
Interpret a transformed formula using points and restrictions.
Explain why a proposed inverse or domain is invalid.
Prerequisites: the three function lessons in this section. Give domains and ranges with each inverse when restrictions matter. A formula alone can leave out essential information about the function.
1. Use a Consistent Set of Checks
Task | Check |
|---|---|
Evaluate f(a) | Is a an allowed input? |
Form f(g(x)) | Is g defined, and does its output enter f's domain? |
Find f⁻¹ | Is f one-to-one on its stated domain? |
Transform a graph | Where does a known point move? |
Find a range | Which outputs occur on the specified domain? |
A denominator, square root, logarithm, or explicit contextual restriction can change which inputs are valid. Keep these conditions during simplification. Two formulas that agree on many inputs need not define the same function if their domains differ.
2. A Function and Its Inverse
Worked Example: A Restricted Quadratic
Let f(x) = (x + 2)² − 3 with domain x ≥ −2. The vertex is (−2, −3), and the range is [−3, ∞).
To invert, write y + 3 = (x + 2)². Since the original x + 2 is nonnegative, take x + 2 = √(y + 3).
Thus f⁻¹(x) = −2 + √(x + 3), with domain [−3, ∞) and range [−2, ∞).
Check f⁻¹(f(x)) = −2 + √((x + 2)²) = −2 + |x + 2| = x on the original domain. The domain restriction justifies the last step.
3. Explain a Failed Shortcut
If a learner writes the inverse of x² as ±√x, they have described a two-valued relation for positive inputs, not a function. Choose a one-to-one branch of the original function first. The selected branch decides the sign of the inverse.
Common Mistakes
An inverse reverses the input-output pairing; a reciprocal divides 1 by the output.
A range statement needs the original domain, especially for quadratics.
The horizontal shift in f(2x − 6) is found by writing f(2(x − 3)), not by reading −6 alone.
4. Mixed Practice
1. Find the domain of √(x − 2)/(x − 5).
2. Find the range of f(x) = (x − 1)² on −1 ≤ x ≤ 2.
3. Let f(x) = 2x − 1 and g(x) = x² + 3. Find f(g(x)) and g(f(x)).
4. Find the inverse of f(x) = 4 − 2x, with all real inputs.
5. The point (8, −2) lies on f. Find its image on g(x) = 3f(2(x − 1)) + 4.
6. Find the domain and range of h(x) = √(x + 4) − 2.
7. Find the inverse of q(x) = x² with domain x ≤ 0, and evaluate q⁻¹(9).
Worked Solutions
1. Require x ≥ 2 and x ≠ 5, so the domain is [2, 5) ∪ (5, ∞).
2. The vertex at x = 1 gives minimum 0. The endpoint values are 4 and 1, so the maximum is 4. The range is [0, 4].
3. f(g(x)) = 2(x² + 3) − 1 = 2x² + 5. Also, g(f(x)) = (2x − 1)² + 3 = 4x² − 4x + 4.
4. From y = 4 − 2x, solve x = (4 − y)/2. Hence f⁻¹(x) = (4 − x)/2, with domain and range both all real numbers.
5. Solve 2(x − 1) = 8 to get x = 5. The new output is 3(−2) + 4 = −2. The image is (5, −2).
6. The domain is [−4, ∞), and the range is [−2, ∞). The square root's starting point becomes (−4, −2).
7. The inverse is q⁻¹(x) = −√x for x ≥ 0. Therefore q⁻¹(9) = −3, which lies in the original function's nonpositive domain.
5. Summary
- Track domains through every function operation.
- Use composition to verify an inverse on the appropriate domain.
- Map known points to check transformations.
- State restrictions as part of the final mathematical answer.