If f(x) = 2x^2 - 3x + 1, what is the value of f(-2)?
- 3
- 15 Correct answer.
- 11
- -1
Explanation
Substitute x = -2 into the rule: 2(-2)^2 - 3(-2) + 1 = 8 + 6 + 1 = 15. The square is applied before multiplication, and subtracting a negative product adds.
Math 30-1 practice questions
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What is the domain of f(x) = sqrt(x - 4)?
- x ≤ 4
- x ≥ -4
- all real numbers
- x ≥ 4 Correct answer.
Explanation
The expression under a square root must be nonnegative, so x - 4 ≥ 0. Solving gives x ≥ 4.
Which relation represents y as a function of x?
- x^2 + y^2 = 9
- y^2 = x
- x = y^2
- y = 2x + 1 Correct answer.
Explanation
A relation is a function if each input x has exactly one output y. The equation y = 2x + 1 is linear, while the others can give two y-values for some x-values.
The graph of y = f(x - 3) + 2 is obtained by translating y = f(x) in which way?
- 3 units right and 2 units up Correct answer.
- 3 units left and 2 units up
- 3 units left and 2 units down
- 3 units right and 2 units down
Explanation
The expression x - 3 inside the function shifts the graph 3 units right. The +2 outside the function shifts the graph 2 units up.
For a one-to-one function f, if f(3) = 7, what is f^{-1}(7)?
- 1/7
- 3 Correct answer.
- -3
- 7
Explanation
An inverse function swaps the input and output of the original function. Since f sends 3 to 7, f^{-1} sends 7 back to 3.
If f(x) = x + 2 and g(x) = 3x, what is f(g(2))?
- 6
- 8 Correct answer.
- 5
- 12
Explanation
Evaluate the inner function first: g(2) = 3(2) = 6. Then f(6) = 6 + 2 = 8.
What is the range of y = x^2 - 4?
- y ≥ -4 Correct answer.
- y ≥ 0
- y ≤ -4
- all real numbers
Explanation
The graph is a parabola opening upward with vertex (0, -4). The vertex gives the minimum y-value, so the range is y ≥ -4.
The function f(x) = x^3 - x is classified as which type?
- odd Correct answer.
- both even and odd
- neither even nor odd
- even
Explanation
Compute f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x). Since f(-x) = -f(x), the function is odd.
If the point (2, 5) lies on the graph of y = f(x), which point lies on the graph of y = f(2x - 6)?
- (4, 5) Correct answer.
- (5, 5)
- (-4, 5)
- (1, 5)
Explanation
The input to f must equal the original x-value 2, so solve 2x - 6 = 2. This gives 2x = 8 and x = 4, so the point is (4, 5).
If f(x) = (x - 3)/2, what is f^{-1}(x)?
- 2x + 3 Correct answer.
- (x - 3)/2
- (x + 3)/2
- 2x - 3
Explanation
Let y = (x - 3)/2, then swap x and y to get x = (y - 3)/2. Solving for y gives y = 2x + 3, so f^{-1}(x) = 2x + 3.
What is the domain of f(x) = (x + 1)/(x^2 - 9)?
- x ≠ ±3 Correct answer.
- x ≠ -3
- all real numbers
- x ≠ 3
Explanation
The denominator factors as (x - 3)(x + 3). A rational expression is undefined where its denominator is zero, so x cannot equal 3 or -3.
If f(x) = sqrt(x) and g(x) = x - 4, what is the domain of f(g(x))?
- all real numbers
- x ≤ 4
- x ≥ 0
- x ≥ 4 Correct answer.
Explanation
The composition is f(g(x)) = sqrt(x - 4). The radicand must be nonnegative, so x - 4 ≥ 0 and x ≥ 4.
What are the vertical and horizontal asymptotes of y = (2x + 1)/(x - 3)?
- x = 3, y = 2 Correct answer.
- x = 2, y = 3
- x = -3, y = 2
- x = 3, y = 0
Explanation
The denominator is zero at x = 3, giving the vertical asymptote x = 3. The numerator and denominator have equal degree, so the horizontal asymptote is the ratio of leading coefficients, y = 2.
What is the solution to sqrt(2x + 3) = x?
- x = -1
- x = 3 or x = -1
- x = 3 Correct answer.
- no solution
Explanation
Squaring both sides gives 2x + 3 = x^2, so x^2 - 2x - 3 = 0 and (x - 3)(x + 1) = 0. Checking in the original equation, x = 3 works but x = -1 is extraneous.
What are the vertex and range of y = |x - 2| + 3?
- (2, 3), y ≤ 3
- (-2, 3), y ≥ 3
- (2, -3), y ≥ -3
- (2, 3), y ≥ 3 Correct answer.
Explanation
The expression inside the absolute value is zero at x = 2, giving the minimum y-value 3, so the vertex is (2, 3). The graph opens upward, so the range is y ≥ 3.
For f(x) = x^2 when x < 0 and f(x) = 2x + 1 when x ≥ 0, what is f(-3) + f(2)?
- 9
- 14 Correct answer.
- 11
- 13
Explanation
Because -3 < 0, use f(x) = x^2, so f(-3) = 9. Because 2 ≥ 0, use f(x) = 2x + 1, so f(2) = 5. The sum is 14.
If f(x) = x^2 and g(x) = x + 1, what is (f ∘ g)(x) - (g ∘ f)(x)?
- 2x Correct answer.
- 2x^2 + 2x
- 2x + 1
- 0
Explanation
Compute (f ∘ g)(x) = (x + 1)^2 = x^2 + 2x + 1 and (g ∘ f)(x) = x^2 + 1. Subtracting gives 2x.
If f(x) = 3^x, what is f^{-1}(x)?
- log_x 3
- 3x
- x^3
- log_3 x Correct answer.
Explanation
Let y = 3^x and swap variables to get x = 3^y. Rewriting in logarithmic form gives y = log_3 x.
What is the domain of f(x) = log_2(x - 5)?
- all real numbers
- x > 5 Correct answer.
- x ≥ 5
- x > 0
Explanation
The argument of a logarithm must be positive, so x - 5 > 0. Solving gives x > 5.
If f(x) = x^2 + 1 for x ≥ 0, what is f^{-1}(x)?
- x^2 - 1
- sqrt(x - 1) Correct answer.
- ±sqrt(x - 1)
- sqrt(x) - 1
Explanation
Let y = x^2 + 1 and solve for x: x^2 = y - 1, so x = sqrt(y - 1) because the domain x ≥ 0 selects the positive root. Thus f^{-1}(x) = sqrt(x - 1).
For f(x) = (x^2 - 4)/(x^2 - x - 2), where is the hole and where is the vertical asymptote?
- hole x = -1, VA x = 2
- hole x = -2, VA x = 2
- VA x = 2 and x = -1
- hole x = 2, VA x = -1 Correct answer.
Explanation
Factor to get (x - 2)(x + 2) over (x - 2)(x + 1). The common factor x - 2 creates a hole at x = 2, while the remaining denominator x + 1 gives a vertical asymptote at x = -1.
If f(x) = (2x + 1)/(x - 3), what is f^{-1}(x)?
- (3x + 1)/(x - 2) Correct answer.
- (x + 3)/(2x + 1)
- (2x - 1)/(x + 3)
- (3x - 1)/(x + 2)
Explanation
Set y = (2x + 1)/(x - 3) and swap x and y: x = (2y + 1)/(y - 3). Solving gives x(y - 3) = 2y + 1, then y(x - 2) = 3x + 1, so y = (3x + 1)/(x - 2).
If f(x) = sqrt(x - 1) and g(x) = 1/(x - 4), what is the domain of f(g(x))?
- x ≥ 1, x ≠ 4
- all real numbers except 4
- (4, 5] Correct answer.
- x < 4 or x ≥ 5
Explanation
The composition is sqrt(1/(x - 4) - 1), and g requires x ≠ 4. The radicand simplifies to (5 - x)/(x - 4), which is nonnegative only for 4 < x ≤ 5, so the domain is (4, 5].
The graph of y = 1/x is reflected in the x-axis, vertically stretched by a factor of 2, translated 2 units left and 3 units up. Which equation represents the transformed graph?
- y = -2/(x + 2) + 3 Correct answer.
- y = -2/(x + 2) - 3
- y = -2/(x - 2) + 3
- y = 2/(x + 2) + 3
Explanation
Reflecting in the x-axis and stretching vertically by 2 gives -2/x. Translating 2 units left replaces x with x + 2, and translating 3 units up adds 3, giving y = -2/(x + 2) + 3.
What is the solution to (x - 1)/(x + 2) ≥ 0?
- -2 < x ≤ 1
- x ≤ -2 or x ≥ 1
- x ≥ 1
- x < -2 or x ≥ 1 Correct answer.
Explanation
The critical values are x = 1, where the numerator is zero, and x = -2, where the denominator is zero. A sign chart shows the expression is positive for x < -2, negative for -2 < x < 1, and nonnegative for x ≥ 1; x = -2 is excluded.
For the function f(x) = 2x^2 - 3x + 1, what is the value of f(-2)?
- 9
- -1
- 15 Correct answer.
- 3
Explanation
Substitute x = -2 into every appearance of x: f(-2) = 2(-2)^2 - 3(-2) + 1. Since (-2)^2 = 4, this gives 2(4) + 6 + 1 = 15. Squaring must be done before multiplying by 2, and subtracting -6 is the same as adding 6.
What is the domain of the function f(x) = sqrt(3 - x)?
- x ≥ 3
- x ≤ -3
- x ≤ 3 Correct answer.
- x ≥ -3
Explanation
The radicand under an even root must be non-negative, so 3 - x ≥ 0. Rearranging this inequality requires dividing by -1, which reverses the inequality sign, giving x ≤ 3.
What is the range of the function y = 2^x + 3?
- (3, ∞) Correct answer.
- (0, ∞)
- (-∞, 3)
- [3, ∞)
Explanation
For every real x, the value of 2^x is positive, so 2^x > 0 and therefore 2^x + 3 > 3. The value 3 is never reached because 2^x never equals 0, so the range is the open interval (3, ∞).
Which set of ordered pairs represents a function?
- {(1, 2), (1, 3), (2, 4), (3, 5)}
- {(1, 2), (2, 3), (3, 4), (4, 5)} Correct answer.
- {(1, 2), (2, 2), (3, 3), (2, 4)}
- {(-1, 0), (0, 1), (-1, 2), (1, 3)}
Explanation
A relation is a function exactly when no input value is paired with two different output values. In only one set does each first coordinate appear once, so every input has a single output.
If f is a one-to-one function and f(3) = 7, what is the value of f^(-1)(7)?
- 3 Correct answer.
- 7
- 1/7
- -3
Explanation
For a one-to-one function, the statement f(a) = b is equivalent to f^(-1)(b) = a. Since f sends 3 to 7, the inverse function sends 7 back to 3.
The point (-2, 6) lies on the graph of y = f(x). Which point must lie on the graph of y = f(-x)?
- (-2, -6)
- (-2, 6)
- (2, -6)
- (2, 6) Correct answer.
Explanation
Replacing x with -x reflects the graph in the y-axis, which changes the sign of each x-coordinate while keeping the y-coordinate the same. Applying this to (-2, 6) gives the point (2, 6).
What is the complete solution of the equation |x - 4| = 7?
- x = 11 only
- x = 3 or x = 11
- x = -3 or x = 11 Correct answer.
- x = -11 or x = 3
Explanation
An absolute value equation splits into two linear cases: x - 4 = 7 or x - 4 = -7. Solving these gives x = 11 and x = -3, and both values check in the original equation.
What is the y-intercept of the graph of f(x) = 2(3)^x + 1?
- (0, 6)
- (0, 2)
- (0, 1)
- (0, 3) Correct answer.
Explanation
The y-intercept occurs where x = 0. Since 3^0 = 1, f(0) = 2(1) + 1 = 3, so the graph crosses the y-axis at (0, 3).
What is the domain of f(x) = sqrt(x + 3)/(x - 1)?
- x > -3
- x ≥ -3
- x > -3, x ≠ 1
- x ≥ -3, x ≠ 1 Correct answer.
Explanation
Two restrictions apply at once: the radicand requires x + 3 ≥ 0, giving x ≥ -3, and the denominator requires x - 1 ≠ 0. Since the radical is in the numerator, x + 3 = 0 is allowed, so the endpoint -3 is included.
If f(x) = (x + 2)^2 - 5 for x ≥ -2, what is f^(-1)(x)?
- sqrt(x + 5) + 2
- -sqrt(x + 5) - 2
- sqrt(x + 5) - 2 Correct answer.
- sqrt(x - 5) - 2
Explanation
Write y = (x + 2)^2 - 5, then isolate x: y + 5 = (x + 2)^2, so x + 2 = ±sqrt(y + 5). The restriction x ≥ -2 means x + 2 ≥ 0, so the principal root is used, giving x = sqrt(y + 5) - 2.
A rational function has a vertical asymptote at x = 2, a horizontal asymptote at y = -1, and an x-intercept of 3. Which equation could represent this function?
- y = (x - 3)/(x - 2)
- y = -(x - 3)/(x + 2)
- y = -(x + 3)/(x - 2)
- y = -(x - 3)/(x - 2) Correct answer.
Explanation
The denominator must be zero at x = 2, giving the factor (x - 2), and the numerator must be zero at x = 3, giving the factor (x - 3). For the horizontal asymptote, the ratio of leading coefficients must equal -1, which requires a negative sign on the numerator.
A population of 1200 bacteria doubles every 5 years. Which equation models the population P after t years?
- P = 1200(2)^t + 5
- P = 1200(5)^(t/2)
- P = 1200(2)^(t/5) Correct answer.
- P = 1200(2)^(5t)
Explanation
Exponential growth is modelled by P = P0(b)^(t/d), where d is the doubling time. With an initial value of 1200, a base of 2, and a doubling period of 5 years, the exponent must be t/5.
What is the range of f(x) = sqrt(x - 3) + 2?
- [0, ∞)
- [3, ∞)
- (-∞, 2]
- [2, ∞) Correct answer.
Explanation
A square root always produces a non-negative output, so sqrt(x - 3) ≥ 0 and therefore f(x) ≥ 2. The minimum value 2 is reached at x = 3, so the range is [2, ∞).
What is the solution set of the inequality |2x - 1| < 5?
- x > 3
- x < -2 or x > 3
- -3 < x < 2
- -2 < x < 3 Correct answer.
Explanation
An absolute value less than a positive number becomes a compound inequality: -5 < 2x - 1 < 5. Adding 1 to all parts gives -4 < 2x < 6, and dividing by 2 gives -2 < x < 3.
For the piecewise function f(x) = x^2 + 1 when x ≤ 0, and f(x) = 2x - 3 when x > 0, what are all values of x for which f(x) = 5?
- x = 4 only
- x = -2 only
- x = -2 or x = 4 Correct answer.
- x = 2 or x = 4
Explanation
Solve each branch separately and check the branch condition. For x ≤ 0, x^2 + 1 = 5 gives x = ±2, but only x = -2 belongs to that branch; for x > 0, 2x - 3 = 5 gives x = 4, which satisfies x > 0.
The graph of y = 2f(-x) + 4 is obtained from the graph of y = f(x) by which combination of transformations?
- Reflection in the y-axis, vertical stretch by a factor of 2, translation 4 units down
- Reflection in the y-axis, vertical stretch by a factor of 2, translation 4 units up Correct answer.
- Reflection in the y-axis, horizontal stretch by a factor of 2, translation 4 units up
- Reflection in the x-axis, vertical stretch by a factor of 2, translation 4 units up
Explanation
Inside the function, x has been replaced by -x, which reflects the graph in the y-axis, and the factor 2 outside the function stretches the graph vertically by a factor of 2. The +4 outside the function translates the graph 4 units upward.
If f(x) = 1/x and g(x) = sqrt(x - 2), what are (f ∘ g)(x) and its domain?
- 1/sqrt(x - 2), domain x ≥ 2
- 1/sqrt(x - 2), domain x > 0
- 1/sqrt(x - 2), domain x > 2 Correct answer.
- sqrt(x - 2), domain x ≥ 2
Explanation
Composing gives (f ∘ g)(x) = f(g(x)) = 1/sqrt(x - 2). The radicand requires x ≥ 2, but the composition also places the radical in a denominator, so sqrt(x - 2) cannot be 0, leaving x > 2.
What is the solution of the equation log base 2 of (x - 1) equals 3?
- x = 7
- x = 9 Correct answer.
- x = 6
- x = 8
Explanation
Rewrite the logarithmic equation in exponential form: x - 1 = 2^3 = 8. Adding 1 to both sides gives x = 9, and this value is in the domain because 9 - 1 = 8 > 0.
What is the solution of 4^(2x - 1) = 8^(x + 1)?
- x = 5 Correct answer.
- x = 4
- x = 1
- x = 2
Explanation
Write both sides with base 2: 4^(2x - 1) = 2^(4x - 2) and 8^(x + 1) = 2^(3x + 3). Equating exponents gives 4x - 2 = 3x + 3, so x = 5.
If f(x) = x^2 - 4 and g(x) = x + 2, what are (f/g)(x) and its domain?
- x + 2, x ≠ -2
- x - 2, x ≠ ±2
- x - 2, x ≠ -2 Correct answer.
- x - 2, x ≠ 2
Explanation
Factor the numerator as (x - 2)(x + 2) and divide by (x + 2), which leaves x - 2 for x ≠ -2. The excluded value comes from the original denominator g(x) = 0, and that restriction remains even though the factor cancels.
If f(x) = 3x - 2 and g(x) = sqrt(x + 1), what is the value of (f ∘ g)^(-1)(4)?
- 1
- 3 Correct answer.
- 4
- 9
Explanation
The equation (f ∘ g)^(-1)(4) = x is equivalent to (f ∘ g)(x) = 4, so solve 3sqrt(x + 1) - 2 = 4. This gives sqrt(x + 1) = 2, then x + 1 = 4, so x = 3.
What is the equation of the oblique asymptote of the graph of y = (x^2 - 1)/(x - 2)?
- y = 2x + 1
- y = x - 2
- y = x + 2 Correct answer.
- y = x
Explanation
Because the numerator has degree one higher than the denominator, divide to get x^2 - 1 = (x - 2)(x + 2) + 3. The quotient x + 2 defines the oblique asymptote, while the remainder term 3/(x - 2) tends to 0 for large |x|.
What are all real solutions of |x^2 - 4| = 3x?
- x = -4, -1, 1, or 4
- x = -1 or x = 4
- x = 1 or x = 4 Correct answer.
- x = 4 only
Explanation
Since the right side must be non-negative, x ≥ 0. Case 1: x^2 - 4 = 3x gives x = 4 or x = -1, and -1 is rejected. Case 2: x^2 - 4 = -3x gives x = 1 or x = -4, and -4 is rejected, leaving x = 1 and x = 4.
The point (3, -2) lies on the graph of y = f(x). Which point lies on the graph of y = -2f((x + 1)/3)?
- (3, 4)
- (9, 4)
- (8, 4) Correct answer.
- (8, -4)
Explanation
The input to f must equal 3, so solve (x + 1)/3 = 3, giving x = 8. The output is then -2 times the original function value: -2(-2) = 4, so the image point is (8, 4).
What is the domain of f(x) = sqrt(x/(x - 2))?
- x ≤ 0 or x ≥ 2
- 0 ≤ x < 2
- x ≤ 0 or x > 2 Correct answer.
- x > 2
Explanation
The radicand must satisfy x/(x - 2) ≥ 0 with x ≠ 2. Testing the intervals cut off by the critical values 0 and 2 shows the quotient is non-negative when x ≤ 0 or x > 2, with x = 2 excluded because the denominator would be zero.