Concept of a function: domain, range, inverse notation 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.

Determine the largest possible domain of f(x) = √(x - 4).

• x > 4
• x ≥ 4
• x ≤ 4
• x ≥ -4

Answer: x ≥ 4

The expression under the square root must be non-negative: x - 4 ≥ 0, so x ≥ 4. The value x = 4 is included because √0 = 0 is defined.

Determine the largest possible domain of g(x) = 1/(x + 3).

• all real x
• x ≠ 3
• x > -3
• x ≠ -3

Answer: x ≠ -3

Division by zero is undefined, so we need x + 3 ≠ 0, i.e. x ≠ -3. A common sign error gives x ≠ 3.

State the range of f(x) = x² + 2, where x ∈ ℝ.

• y ≥ 2
• y > 2
• y ≥ 0
• y ∈ ℝ

Answer: y ≥ 2

Since x² ≥ 0 for all real x, the minimum value of x² + 2 is 2, attained at x = 0; hence the range is y ≥ 2.

Given f(x) = 3x - 6, determine f⁻¹(x).

• f⁻¹(x) = (x - 6)/3
• f⁻¹(x) = 1/(3x - 6)
• f⁻¹(x) = (x + 6)/3
• f⁻¹(x) = 3x + 6

Answer: f⁻¹(x) = (x + 6)/3

Set y = 3x - 6, swap variables and solve: x = 3y - 6 gives y = (x + 6)/3. Note that f⁻¹(x) is not the same as 1/f(x).

Which condition must a function f satisfy so that its inverse function f⁻¹ exists?

• f is one-to-one on its domain
• f is continuous on its domain
• f passes the vertical line test
• f is defined for all real numbers

Answer: f is one-to-one on its domain

An inverse exists only if f is one-to-one, so that each output comes from exactly one input; graphically f must pass the horizontal line test.