AP Calculus AB Flashcards: Candidates Test

What this deck covers

This deck focuses on Candidates Test, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus AB.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What are the critical points of $f(x) = x^2 - 9x + 20$?

Answer: Critical points: $x = \frac{9}{2}$. Set $f'(x)=2x-9=0$ and solve for $x$.

Flashcard 2: Find $f(x)$ at $x = 3$ for $f(x) = x^3 - 6x^2 + 9x + 1$.

Answer: $f(3) = 1$. Substitute $x=3$ into the function.

Flashcard 3: If $f'(x)$ changes sign at $x = c$, what is $x = c$?

Answer: A critical point. Sign changes in derivative indicate critical points.

Flashcard 4: What is a critical point?

Answer: Where $f'(x) = 0$ or $f'(x)$ is undefined. Points where derivative equals zero or doesn't exist.

Flashcard 5: Which function value determines the absolute maximum?

Answer: The largest value among evaluated points. Highest function output among all tested candidates.

Flashcard 6: What must be determined after finding the derivative in the Candidates Test?

Answer: Critical points where the derivative is zero or undefined. These are locations where extrema can occur on the interval.

Flashcard 7: Find the derivative of $f(x) = x^3 - 3x^2 + 4$.

Answer: $f'(x) = 3x^2 - 6x$. Apply power rule: $\frac{d}{dx}[x^3]=3x^2$, etc.

Flashcard 8: Evaluate $f'(x) = 6x - 6$ for critical points.

Answer: Critical point: $x = 1$. Set $6x-6=0$ and solve for $x$.

Flashcard 9: Find $f(x)$ at $x = 5$ for $f(x) = x^2 - 5x$.

Answer: $f(5) = 0$. Substitution shows $x=5$ is a root of the function.

Flashcard 10: Determine the endpoints for the interval $[a, b] = [-2, 5]$.

Answer: Endpoints are $x = -2$ and $x = 5$. These are the boundary values of the given interval.

Flashcard 11: Evaluate $f'(x) = 4x - 4$ for critical points.

Answer: Critical point: $x = 1$. Set $4x-4=0$ and solve for $x$.

Flashcard 12: Find $f(x)$ at $x = 0$ for $f(x) = x^4 - 2x^2$.

Answer: $f(0) = 0$. Direct substitution of $x=0$ into the function.

Flashcard 13: Identify the type of interval required for the Candidates Test.

Answer: A closed interval $[a, b]$. Ensures function is continuous and extrema exist by EVT.

Flashcard 14: State the condition for a point to be a candidate for extrema.

Answer: Point must be a critical point or endpoint. Only these points can produce absolute extrema on closed intervals.

Flashcard 15: State the first step in the Candidates Test.

Answer: Find the derivative of the function. Need derivative to locate critical points where extrema may occur.

Flashcard 16: What is the Candidates Test used for in calculus?

Answer: Determining the absolute extrema on a closed interval. Finds global max/min on closed intervals by testing all candidate points.

Flashcard 17: Find $f(1)$ for $f(x) = x^2 - 2x + 1$.

Answer: $f(1) = 0$. This is a perfect square: $(x-1)^2=0$ at $x=1$.

Flashcard 18: Identify the formula for finding critical points.

Answer: Set $f'(x) = 0$ and solve for $x$. Standard method to find where slope equals zero.

Flashcard 19: Find $f(x)$ at $x = 5$ for $f(x) = x^2 - 5x$.

Answer: $f(5) = 0$. Substitution shows $x=5$ is a root of the function.

Flashcard 20: What is $f(x)$ at $x = 0$ for $f(x) = \frac{1}{x} + x$ on $[1, 2]$?

Answer: Not applicable; $x = 0$ is not in $[1, 2]$. Point outside domain cannot be evaluated.

Flashcard 21: Determine $f(x)$ at $x=2$ for $f(x) = x^3 - 3x^2 + 4x$.

Answer: $f(2) = 4$. Direct substitution of $x=2$ into the cubic function.

Flashcard 22: Determine the endpoints for the interval $[a, b] = [-2, 5]$.

Answer: Endpoints are $x = -2$ and $x = 5$. These are the boundary values of the given interval.

Flashcard 23: Find $f(x)$ at $x = 0$ for $f(x) = x^4 - 2x^2$.

Answer: $f(0) = 0$. Direct substitution of $x=0$ into the function.

Flashcard 24: Which function value determines the absolute minimum?

Answer: The smallest value among evaluated points. Lowest function output among all tested candidates.

Flashcard 25: Evaluate $f(x)$ at endpoints for $f(x) = \frac{1}{x}$ on $[1, 3]$.

Answer: $f(1) = 1$, $f(3) = \frac{1}{3}$. Substitute endpoint values into $f(x)=\frac{1}{x}$.

Flashcard 26: Which values are evaluated in the Candidates Test?

Answer: Endpoints and critical points of the interval. All possible locations where absolute extrema can occur.

Flashcard 27: What is the final step in the Candidates Test?

Answer: Compare function values at critical points and endpoints. Determines which candidate gives the absolute maximum/minimum.

Flashcard 28: Find $f(1)$ for $f(x) = x^2 - 2x + 1$.

Answer: $f(1) = 0$. This is a perfect square: $(x-1)^2=0$ at $x=1$.

Flashcard 29: Evaluate $f'(x) = 3x^2 - 12x + 9$ for critical points.

Answer: Critical points: $x = 1, x = 3$. Solve $3x^2-12x+9=0$ to get $x=1,3$.

Flashcard 30: Evaluate $f'(x) = 4x - 4$ for critical points.

Answer: Critical point: $x = 1$. Set $4x-4=0$ and solve for $x$.