AP Calculus AB Flashcards: Candidates Test

Study Candidates Test in AP Calculus AB with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.

AP Calculus AB

Candidates Test

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QUESTION

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

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ANSWER

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

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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.

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Flashcard 1: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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 9: If $f'(x)$ changes sign at $x = c$ , what is $x = c$ ?

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

Flashcard 10: Evaluate $f(x)$ at endpoints for $f(x) = x^2 - 2x$ on $[0, 3]$ .

Answer: $f(0) = 0$ , $f(3) = 3$ . Substitute boundary values into the function.

Flashcard 11: 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 12: 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 13: 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 14: Which function value determines the absolute maximum?

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

Flashcard 15: 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 16: 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 17: 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 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: 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 20: Evaluate $f'(x) = 4x - 4$ for critical points.

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

Flashcard 21: 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 22: Evaluate $f(x)$ at endpoints for $f(x) = x^2 - 2x$ on $[0, 3]$ .

Answer: $f(0) = 0$ , $f(3) = 3$ . Substitute boundary values into the function.

Flashcard 23: 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 24: 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 25: 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 26: 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 27: 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 28: 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 29: 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 30: 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 31: 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 32: If $f'(x)$ changes sign at $x = c$ , what is $x = c$ ?

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

Flashcard 33: Find $f(x)$ at $x = 2$ for $f(x) = x^2 - 4x + 4$ .

Answer: $f(2) = 0$ . Direct substitution shows this is a perfect square.

Flashcard 34: Evaluate $f'(x) = 6x - 6$ for critical points.

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

Flashcard 35: 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 36: 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 37: Evaluate $f'(x) = 6x - 6$ for critical points.

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

Flashcard 38: 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 39: 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 40: 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 41: Find $f(x)$ at $x = 2$ for $f(x) = x^2 - 4x + 4$ .

Answer: $f(2) = 0$ . Direct substitution shows this is a perfect square.

Flashcard 42: Which function value determines the absolute minimum?

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

Flashcard 43: Evaluate $f'(x) = 4x - 4$ for critical points.

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

Flashcard 44: 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 45: 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 46: What is the derivative of $f(x) = x^3 - 6x^2 + 9x + 1$ ?

Answer: $f'(x) = 3x^2 - 12x + 9$ . Apply power rule to each term.

Flashcard 47: Which function value determines the absolute maximum?

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

Flashcard 48: 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 49: 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 50: 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 51: 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 52: What are endpoints in the context of the Candidates Test?

Answer: The boundary values of the closed interval. The left and right limits of the domain interval.

Flashcard 53: Which function value determines the absolute minimum?

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

Flashcard 54: 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 55: What are endpoints in the context of the Candidates Test?

Answer: The boundary values of the closed interval. The left and right limits of the domain interval.

Flashcard 56: 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 57: Find $f(1)$ for $f(x) = x^2 - 2x + 1$ .

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

AP Calculus AB Flashcards: Candidates Test

Candidates Test

What this deck covers

How to use these flashcards

All flashcards

Flashcard 1: Find f(x)f(x)f(x) at x=5x = 5x=5 for f(x)=x2−5xf(x) = x^2 - 5xf(x)=x2−5x.

Flashcard 2: What is the Candidates Test used for in calculus?

Flashcard 3: State the condition for a point to be a candidate for extrema.

Flashcard 4: Determine the endpoints for the interval [a,b]=[−2,5][a, b] = [-2, 5][a,b]=[−2,5].

Flashcard 5: Determine f(x)f(x)f(x) at x=2x=2x=2 for f(x)=x3−3x2+4xf(x) = x^3 - 3x^2 + 4xf(x)=x3−3x2+4x.

Flashcard 6: What are the critical points of f(x)=x2−9x+20f(x) = x^2 - 9x + 20f(x)=x2−9x+20?

Flashcard 7: Identify the type of interval required for the Candidates Test.

Flashcard 8: Find f(x)f(x)f(x) at x=0x = 0x=0 for f(x)=x4−2x2f(x) = x^4 - 2x^2f(x)=x4−2x2.

Flashcard 9: If f′(x)f'(x)f′(x) changes sign at x=cx = cx=c, what is x=cx = cx=c?

Flashcard 10: Evaluate f(x)f(x)f(x) at endpoints for f(x)=x2−2xf(x) = x^2 - 2xf(x)=x2−2x on [0,3][0, 3][0,3].

Flashcard 11: What must be determined after finding the derivative in the Candidates Test?

Flashcard 12: Evaluate f(x)f(x)f(x) at endpoints for f(x)=1xf(x) = \frac{1}{x}f(x)=x1​ on [1,3][1, 3][1,3].

Flashcard 13: State the first step in the Candidates Test.

Flashcard 14: Which function value determines the absolute maximum?

Flashcard 15: Find f(1)f(1)f(1) for f(x)=x2−2x+1f(x) = x^2 - 2x + 1f(x)=x2−2x+1.

Flashcard 16: What is the final step in the Candidates Test?

Flashcard 17: Find the derivative of f(x)=x3−3x2+4f(x) = x^3 - 3x^2 + 4f(x)=x3−3x2+4.

Flashcard 18: Identify the formula for finding critical points.

Flashcard 19: What is a critical point?

Flashcard 20: Evaluate f′(x)=4x−4f'(x) = 4x - 4f′(x)=4x−4 for critical points.

Flashcard 21: Determine the endpoints for the interval [a,b]=[−2,5][a, b] = [-2, 5][a,b]=[−2,5].

Flashcard 22: Evaluate f(x)f(x)f(x) at endpoints for f(x)=x2−2xf(x) = x^2 - 2xf(x)=x2−2x on [0,3][0, 3][0,3].