Concept of a function; domain, range; inverse via graph practice questions.

Given f(x) = 2x + 1, find f(3).

• 5
• 6
• 7
• 8

Answer: 7

To evaluate a function, substitute the input for x: f(3) = 2 × 3 + 1 = 7.

For the function g(x) = √(x − 4), state the domain.

• x ≥ 4
• x > 4
• x ≥ 0
• x ≤ 4

Answer: x ≥ 4

The expression under a square root must be ≥ 0, so x − 4 ≥ 0 gives x ≥ 4 (equality allowed since √0 = 0).

The function h is defined by h(x) = 1/(x − 3). State the range of h.

• h(x) ≠ 0
• h(x) ≠ 3
• h(x) > 0
• all real numbers

Answer: h(x) ≠ 0

A reciprocal function never outputs 0, so the range excludes 0 (the horizontal asymptote is y = 0).

A function is a mapping in which each element of the domain is associated with:

• exactly one element of the range
• at least two elements of the range
• no element of the range
• every element of the range

Answer: exactly one element of the range

By definition a function assigns exactly one output to each input; this is what the vertical line test checks graphically.

For f(x) = 2ˣ, determine the domain and range.

• domain x ∈ ℝ, range f(x) > 0
• domain x > 0, range f(x) ∈ ℝ
• domain x ∈ ℝ, range f(x) ≥ 0
• domain x > 0, range f(x) > 0

Answer: domain x ∈ ℝ, range f(x) > 0

An exponential accepts every real input, and 2ˣ is always positive (approaching but never reaching 0), so the range is f(x) > 0.

Given f(x) = (x + 5)/2, find an expression for f⁻¹(x).

• 2x − 5
• 2x + 5
• (x − 5)/2
• 2/(x + 5)

Answer: 2x − 5

Set y = (x + 5)/2, swap and solve: x = (y + 5)/2 → 2x = y + 5 → y = 2x − 5. The distractor (x−5)/2 wrongly keeps the ÷2.