Questions: Inverse Functions

Question 1

If f(x) = 2x + 6, what is f⁻¹(x)?

• A. 1/(2x + 6) — the reciprocal of f(x)
• B. 2x − 6 — subtracting 6 from the original function
• C. (x − 6)/2 — swapping x and y, then solving for y
• D. (x + 6)/2 — a sign error when isolating the variable

Answer: (C)

To find the inverse, swap x and y in y = 2x + 6 to get x = 2y + 6, then solve for y: y = (x − 6)/2. Verify by composition: f(f⁻¹(x)) = 2·((x−6)/2) + 6 = (x−6) + 6 = x. ✓ Option A is the most common mistake — confusing the inverse function f⁻¹ with the reciprocal 1/f(x). These are completely different: f⁻¹ undoes f, while 1/f(x) is just the multiplicative reciprocal. Option D gets the sign wrong when solving 2y = x − 6.

Question 2

Which function does NOT have an inverse over its natural domain?

• A. f(x) = 3x − 1
• B. f(x) = x³
• C. f(x) = x²
• D. f(x) = eˣ

Answer: (C)

f(x) = x² fails the horizontal line test on its natural domain (all real numbers): the horizontal line y = 4 crosses the graph at both x = 2 and x = −2. Since two different inputs produce the same output (f(2) = f(−2) = 4), the function is not one-to-one, and no inverse exists. The other three functions are all strictly monotone over their domains — f(x) = 3x−1 and f(x) = eˣ are strictly increasing, f(x) = x³ is strictly increasing everywhere — so each passes the horizontal line test.

Question 3

The notation f⁻¹(x) means the same thing as [f(x)]⁻¹ = 1/f(x).

• T. True
• F. False

Answer: False

False — this is one of the most common misconceptions in algebra. f⁻¹(x) denotes the inverse function, which undoes what f does: if f(a) = b, then f⁻¹(b) = a. The expression 1/f(x) is the reciprocal, which has no such relationship. For example, if f(x) = 2x, then f⁻¹(x) = x/2 (divides by 2), while 1/f(x) = 1/(2x) (a completely different function). The notation is unfortunate but standard — the exponent −1 on a function name always means 'inverse function,' not 'reciprocal.'

Question 4

The graph of f⁻¹ is the reflection of the graph of f across the line y = x.

• T. True
• F. False

Answer: True

True. Reflecting a graph across the line y = x swaps every point (a, b) to (b, a) — which is exactly the relationship between f and f⁻¹. If the point (a, b) is on the graph of f (meaning f(a) = b), then the point (b, a) is on the graph of f⁻¹ (meaning f⁻¹(b) = a). This geometric fact explains why the inverse can be found algebraically by swapping x and y: swapping coordinates in the equation is the algebraic version of reflecting over y = x.

Question 5

Explain why f(x) = x² does not have an inverse on its natural domain, and describe how restricting the domain to x ≥ 0 fixes this problem.

Model answer: f(x) = x² is not one-to-one on all real numbers because it maps two different inputs to the same output: f(2) = 4 and f(−2) = 4. An inverse function would need to answer 'what input produced 4?' — but there are two valid answers (2 and −2), making the inverse ambiguous and not a function. Restricting the domain to x ≥ 0 removes the negative inputs, keeping only the right half of the parabola. This restricted version is strictly increasing, so each output corresponds to exactly one input. The inverse of x² on x ≥ 0 is √x (with range y ≥ 0), and the two graphs are reflections of each other across y = x.

Domain restriction is a general technique: any non-one-to-one function can be made invertible by selecting a portion of its domain on which it is monotone. For x², x ≥ 0 is the conventional choice, giving the standard square root. The inverse only exists on the restricted domain, and its range is restricted accordingly — this is why √x returns only non-negative values.