Precalculus Practice Tests #1: Functions & Transformations

Overview

blackpenredpen works through all 20 multiple-choice questions in the first of 10 planned precalculus practice tests, focusing on functions and transformations. The solutions cover graph-based domain and range, difference quotients and average rates of change, composition and inverse functions, transformation rules, function properties, and modeling geometry with algebraic formulas.

Key takeaways

For graph-based domain and range questions, endpoint inclusion is determined by filled versus open circles; the first graph has domain [-4, 4) and range [-1, 3].
For a difference quotient, expand f(x + h), subtract f(x) with parentheses, and factor h before canceling; for 3x² - 2x + 1 the result is 6x + 3h - 2.
Domain restrictions come from the expression's operations: even roots require a nonnegative radicand, denominators cannot be zero, and logarithm inputs must be positive.
Transformations inside a function affect horizontal position or scale in the opposite direction: f(2x) compresses horizontally by 1/2, while f(x + 4) shifts left 4.
A function has an inverse on its domain exactly when it is one-to-one, as tested graphically by the horizontal line test; a parabola needs a restricted domain to become one-to-one.
Known ranges can be transformed by applying outside operations to both endpoints; for (1/2)f(2x + 3) - 4 with range f = [-2, 5], the resulting range is [-5, -1.5].

Chapters

0:00 Practice Test Setup and Reading Domain from a Graph

Introduces the first of 10 planned precalculus practice tests; each test contains 20 multiple-choice questions.
For Question 1, reads the graph's x-values from -4 to 4: the closed circle includes -4, while the open circle excludes 4.
Finds the graph's range from -1 to 3, including both endpoints because the graph includes them.

2:20 Question 2: Range of a Reflected Absolute-Value Function

Graphs f(x) = 3 - |x + 2| by shifting |x| left 2 units, reflecting it over the x-axis, and shifting it up 3.
The vertex is (-2, 3), the maximum y-value; the graph continues downward without bound.
Concludes that the range is (-infinity, 3], including the vertex value 3.

5:30 Question 3: Simplifying a Difference Quotient

For f(x) = 3x² - 2x + 1, substitutes x + h into every x to calculate f(x + h).
Expands and subtracts f(x), carefully distributing the minus sign so the constant and x² terms cancel.
Factors h from the numerator and cancels it with the denominator, giving 6x + 3h - 2.

10:23 Question 4: Average Rate of Change on an Interval of Length h

Uses the slope formula [f(b) - f(a)] / (b - a) for f(x) = x³ - 2x on [1, 1 + h].
Expands (1 + h)³ using the coefficients 1, 3, 3, 1 from Pascal's triangle.
After subtracting f(1) and canceling h, obtains the average rate of change 1 + 3h + h².

17:07 Questions 5–6: Function Composition, Domain, and Average Change

Composes f(x) = √(x - 1) with g(x) = 2x + 3 to get f(g(x)) = √(2x + 2).
Applies the even-root restriction 2x + 2 ≥ 0, giving the domain [-1, infinity).
For f(x) = √(x + 1) on [3, 8], evaluates f(3) = 2 and f(8) = 3; the average rate of change is 1/5.

23:40 Question 7: Finding a Rational Function's Inverse

Finds the inverse of f(x) = (2x + 1) / (x - 3) by writing y = f(x) and switching x and y.
Clears the denominator, collects the y-terms, and factors y to isolate the new output.
Gets f⁻¹(x) = (3x + 1) / (x - 2); emphasizes that inverse notation does not mean taking a reciprocal.

26:32 Questions 8–9: Writing and Reading Function Transformations

Transforms |x| with a vertical compression by 1/3, a shift left 4, and a shift down 5 to form (1/3)|x + 4| - 5.
Reads g(x) = -3f(x - 2) + 4 from the inside out: shift right 2, reflect across the x-axis, vertically stretch by 3, then shift up 4.
Distinguishes vertical compression for a positive multiplier between 0 and 1 from vertical stretch for a multiplier greater than 1.

35:26 Question 10: Mapping a Point Through a Function Transformation

Maps the original point (3, -6) through a transformation that reflects the input, doubles the function output, and adds 1.
The input reflection changes x = 3 to x = -3; the output change gives 2(-6) + 1 = -11.
The corresponding point is (-3, -11), illustrating that changes inside f affect x-coordinates while outside changes affect y-coordinates.

39:11 Questions 11–12: Odd Functions and Rational-Root Domains

Tests f(x) = x³ / (x² + 1) by substituting -x; the result is -f(x), so the function is odd.
For √(9 - x²) / (2x + 1), requires 9 - x² ≥ 0 and 2x + 1 ≠ 0.
Combines the restrictions to get the domain [-3, -1/2) union (-1/2, 3].

48:18 Question 13: Modeling a Cone's Volume as a Function

Uses the cone formula V = (1/3)πr²h and defines the base radius as x.
Applies the condition that height is twice the radius, setting h = 2x.
Substitution yields the volume function V(x) = (2π/3)x³.

52:08 Questions 14–15: Asymptotes and Horizontal Compression

For 1/(x + 3) - 2, shifts the parent graph 1/x left 3 and down 2, producing vertical asymptote x = -3 and horizontal asymptote y = -2.
Explains that f(2x) changes the graph horizontally: an original point with x = 1 appears at x = 1/2.
Identifies f(2x) as a horizontal compression by a factor of 1/2; multiplying the input by a value greater than 1 narrows the graph.

1:00:24 Question 16: Transforming a Function's Range

Starts with range f(x) in [-2, 5] and applies the positive output multiplier 1/2, producing [-1, 2.5].
The input 2x + 3 does not restrict the range here because it is a linear expression that can attain every real input to f.
Subtracting 4 gives the transformed range [-5, -1.5] for (1/2)f(2x + 3) - 4.

1:09:27 Question 17: Rectangle Area from a Fixed Perimeter

Models a rectangle with perimeter 20 and one side of length x; opposite sides have the same lengths.
From 2x + 2w = 20, solves for the other side as w = 10 - x.
Writes the area as A(x) = x(10 - x) = 10x - x².

1:13:29 Question 18: Continuous Functions with an Inverse

Checks both continuity and the horizontal line test for functions defined over all real numbers.
Rejects x² - 4 and |x| + 1 because they fail the horizontal line test, and 1/(x - 1) because it is discontinuous at x = 1.
Selects x³ + 2: it is continuous and one-to-one, so it has an inverse.

1:18:08 Question 19: Restricting a Parabola to Find Its Inverse

For f(x) = (x - 2)² + 3, identifies the vertex as (2, 3) and restricts the domain to x ≥ 2 so the function is one-to-one.
Switches x and y, then solves for y using only the positive square root because the chosen branch is the right half of the parabola.
Finds the inverse f⁻¹(x) = √(x - 3) + 2, with inverse domain x ≥ 3.

1:21:53 Question 20: Identifying a Function That Is Not One-to-One

Uses the horizontal line test to identify the graph that does not have an inverse function.
The exponential, shifted reciprocal, and square-root graphs shown pass the test.
Selects the parabola as the non-one-to-one graph because some horizontal lines intersect it twice.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, blackpenredpen.