Functions

1. What is the natural real domain of f(x) = 1/(x - 4)?

Answer A: All real x except x = 4

The denominator x - 4 is zero at x = 4, so that input must be excluded. Every other real input gives exactly one real output.

2. For real x, what is the range of g(x) = x^2 + 5?

Answer B: g(x) >= 5

Since x^2 >= 0, g(x) >= 5. The value 5 is attained at x = 0, and every larger value is produced, so the boundary is included.

3. Find the range of h(x) = 1/x when its domain is x >= 2.

Answer C: 0 < h(x) <= 1/2

The maximum 1/2 occurs at x = 2. As x increases, 1/x remains positive and approaches but never reaches zero, so 0 < h(x) <= 1/2.

4. Which relation does not define y as a function of x?

Answer D: x = y^2 for real y

The relation x = y^2 gives y = +/-sqrt(x) for x > 0, so one x-value has two y-values. Each other relation assigns a unique y to every allowed x.

5. Given f(x) = 2x - 3 and g(x) = x^2, find gf(x).

Answer A: (2x - 3)^2

gf(x) = g(f(x)) = g(2x - 3). Since g squares its input, gf(x) = (2x - 3)^2.

6. Let f(x) = sqrt(x) and g(x) = x - 5. What are fg(x) and its real domain?

Answer B: sqrt(x - 5), x >= 5

fg(x) = f(g(x)) = sqrt(x - 5). Real outputs require x - 5 >= 0, so the composite domain is x >= 5.

7. Which condition ensures that gf can be formed on the whole stated domain of f?

Answer C: The range of f lies within the domain of g

For gf(x) = g(f(x)), every output supplied by f must be accepted by g. Therefore the range of f must lie within the domain of g.

8. Let f(x) = x^2 for -1 <= x <= 3 and g(x) = 1/(x - 4). What is the domain of gf?

Answer D: -1 <= x <= 3, with x = 2 excluded

gf(x) = 1/(x^2 - 4). Within -1 <= x <= 3, the denominator is zero only at x = 2; x = -2 is outside the domain. Thus exclude x = 2.

9. Which function is one-one on the set of all real numbers?

Answer A: f(x) = 3x - 7

The linear function 3x - 7 is strictly increasing, so distinct inputs give distinct outputs. Each other option gives equal outputs at inputs symmetric about a point.

10. The function f(x) = x^2 is restricted to x >= 0. What is f inverse(x)?

Answer B: sqrt(x), x >= 0

From y = x^2 and x >= 0, rearrangement gives x = sqrt(y). Therefore f inverse(x) = sqrt(x), whose domain is the original range x >= 0.

11. If h(x) = 5 - 2x, find h inverse(x).

Answer C: (5 - x)/2

From 2x = 5 - y, x = (5 - y)/2. Swapping variable names gives h inverse(x) = (5 - x)/2.

12. Let f(x) = (x - 1)^2 + 4 for x <= 1. Which is its inverse?

Answer D: 1 - sqrt(x - 4), x >= 4

Rearranging gives x - 1 = -sqrt(y - 4) because x <= 1. Hence f inverse(x) = 1 - sqrt(x - 4), with domain x >= 4.

13. For f(x) = (3x - 5)/(x + 2), what is f inverse(x)?

Answer A: (2x + 5)/(3 - x)

yx + 2y = 3x - 5 gives x(y - 3) = -5 - 2y, so x = (2y + 5)/(3 - y). Swap x and y to obtain the stated inverse.

14. A one-one function satisfies f(2) = 7. What must be true?

Answer B: f inverse(7) = 2

The mapping 2 to 7 under f is reversed by the inverse, so f inverse maps 7 to 2. On their graphs, (2, 7) and (7, 2) are reflected points.

15. The graph of a one-one function is reflected to obtain the graph of its inverse. What is the mirror line?

Answer C: y = x

Reflection in y = x swaps x- and y-coordinates. Thus each point (a, b) on f corresponds to (b, a) on f inverse.

16. A one-one function has domain x > 1 and range y < 4. What are the domain and range of its inverse?

Answer D: Domain x < 4, range y > 1

The original range becomes the inverse domain, so inverse inputs satisfy x < 4. The original domain becomes the inverse range, so inverse outputs satisfy y > 1.

17. How is y = f(x) + 3 obtained from y = f(x)?

Answer A: Translate up 3 units

Adding 3 to every output raises every y-coordinate by 3. The translation vector is (0, 3).

18. How is y = f(x + 4) obtained from y = f(x)?

Answer B: Translate left 4 units

Each old point with x-coordinate p moves to p - 4, so the graph translates 4 units left. The translation vector is (-4, 0).

19. Which transformation maps y = f(x) to y = -f(x)?

Answer C: Reflection in the x-axis

Multiplying all y-values by -1 maps each point (p, q) to (p, -q), which is reflection in the x-axis.

20. Which transformation maps y = f(x) to y = f(-x)?

Answer D: Reflection in the y-axis

Replacing x by -x maps each point (p, q) to (-p, q), which is reflection in the y-axis.

21. How is y = 3f(x) obtained from y = f(x)?

Answer A: Vertical stretch by factor 3

Each point (p, q) maps to (p, 3q), so distances from the x-axis triple. This is a vertical stretch by factor 3.

22. How is y = f(2x) obtained from y = f(x)?

Answer B: Horizontal scale factor 1/2

The new x-coordinate is p/2, so every horizontal distance from the y-axis is halved. The horizontal scale factor is 1/2.

23. The point (6, -4) lies on y = f(x). Which point lies on y = -2f(3x)?

Answer C: (2, 8)

Solve 3x = 6 to obtain x = 2, and multiply -4 by -2 to obtain y = 8. The corresponding point is (2, 8).

24. The point (-3, 4) lies on y = f(x). Which point lies on y = 2f(-(x - 1)) + 5?

Answer D: (4, 13)

From -(x - 1) = -3, x - 1 = 3 and x = 4. The new output is 2(4) + 5 = 13, so the mapped point is (4, 13).