AP Calculus BC Flashcards: Determining Intervals On Increasing Decreasing Functions

What this deck covers

This deck focuses on Determining Intervals On Increasing Decreasing Functions, giving you a quick way to review the definitions, rules, and examples that matter most for AP Calculus BC.

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 is the test to determine if a function is increasing on an interval?

Answer: The first derivative is positive. When $f'(x) > 0$, the function slopes upward.

Flashcard 2: How can you determine the critical points of $f(x) = x^3 - 6x^2$?

Answer: Solve $f'(x) = 0$: $x = 0, 4$. $f'(x) = 3x^2 - 12x = 3x(x-4)$, set equal to zero.

Flashcard 3: Find intervals where $f(x) = x^2 - 8x + 15$ is increasing.

Answer: $x \in (4, \infty)$. $f'(x) = 2x - 8$, positive when $x > 4$.

Flashcard 4: What is the second derivative of $f(x) = x^3 - 3x^2 + 2x$?

Answer: $f''(x) = 6x - 6$. Take derivative of $f'(x) = 3x^2 - 6x + 2$.

Flashcard 5: If $f'(x) < 0$ and $f''(x) > 0$, what can be said about $f(x)$?

Answer: $f(x)$ is decreasing and concave up. Function falls while curving upward like a valley.

Flashcard 6: What does a negative $f'(x)$ indicate about $f(x)$ on an interval?

Answer: $f(x)$ is decreasing. Negative slope means function values are falling.

Flashcard 7: For $f'(x) = -4x$, on which interval is $f(x)$ decreasing?

Answer: $x \in (0, \infty)$. $f'(x) = -4x < 0$ when $x > 0$.

Flashcard 8: What does $f'(x) = (x-1)^2$ imply about intervals of increase/decrease?

Answer: $f(x)$ is non-decreasing. $f'(x) = (x-1)^2 \geq 0$, never decreasing.

Flashcard 9: State the critical point condition for a function $f(x)$.

Answer: $f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.

Flashcard 10: Find intervals where $f(x) = x^2 - 8x + 15$ is decreasing.

Answer: $x \in (-\infty, 4)$. $f'(x) = 2x - 8$, negative when $x < 4$.

Flashcard 11: Find intervals where $f(x) = x^2 - 8x + 15$ is decreasing.

Answer: $x \in (-\infty, 4)$. $f'(x) = 2x - 8$, negative when $x < 4$.

Flashcard 12: For $f'(x) = -4x$, on which interval is $f(x)$ decreasing?

Answer: $x \in (0, \infty)$. $f'(x) = -4x < 0$ when $x > 0$.

Flashcard 13: What does a negative $f'(x)$ indicate about $f(x)$ on an interval?

Answer: $f(x)$ is decreasing. Negative slope means function values are falling.

Flashcard 14: What is the condition for $f(x)$ to be constant on an interval?

Answer: $f'(x) = 0$ for all $x$ in that interval. Zero derivative means no rate of change.

Flashcard 15: What is the test to determine if a function is decreasing on an interval?

Answer: The first derivative is negative. When $f'(x) < 0$, the function slopes downward.

Flashcard 16: What is a necessary condition for a point to be a local extremum?

Answer: $f'(x) = 0$ or $f'(x)$ is undefined. Local extrema occur only at critical points.

Flashcard 17: What does $f'(x) = 0$ at $x = a$ indicate?

Answer: Possible local extremum. Critical points may be local maxima, minima, or inflection points.

Flashcard 18: What does a positive $f'(x)$ indicate about $f(x)$ on an interval?

Answer: $f(x)$ is increasing. Positive slope means function values are rising.

Flashcard 19: What is the first derivative test used for?

Answer: To determine increasing or decreasing intervals. Analyzes sign of $f'(x)$ to find where function rises or falls.

Flashcard 20: How can you determine the critical points of $f(x) = x^3 - 6x^2$?

Answer: Solve $f'(x) = 0$: $x = 0, 4$. $f'(x) = 3x^2 - 12x = 3x(x-4)$, set equal to zero.

Flashcard 21: State the critical point condition for a function $f(x)$.

Answer: $f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.

Flashcard 22: What is the relationship between critical points and intervals of increase/decrease?

Answer: Critical points separate intervals. They divide the domain into regions with consistent behavior.

Flashcard 23: For $f'(x) = 4x^3 - 12x$, find the critical points.

Answer: $x = 0, \pm \sqrt{3}$. $f'(x) = 4x(x^2 - 3)$, set equal to zero.

Flashcard 24: What is the derivative of $f(x) = x^3 - 3x^2$?

Answer: $f'(x) = 3x^2 - 6x$. Power rule: bring down exponent and reduce by 1.

Flashcard 25: What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$?

Answer: $f(x)$ is strictly increasing on $(a, b)$. Positive derivative throughout means consistently rising.

Flashcard 26: What is a necessary condition for a point to be a local extremum?

Answer: $f'(x) = 0$ or $f'(x)$ is undefined. Local extrema occur only at critical points.

Flashcard 27: What is the second derivative test used for?

Answer: To determine concavity. Analyzes $f''(x)$ to determine curve bending direction.

Flashcard 28: What does $f''(x) < 0$ indicate about $f(x)$?

Answer: $f(x)$ is concave down. Negative second derivative means curve bends downward.

Flashcard 29: What does $f'(x) = 0$ but no sign change indicate at a critical point?

Answer: No local extremum. No sign change means no local maximum or minimum.

Flashcard 30: What is the test to determine if a function is decreasing on an interval?

Answer: The first derivative is negative. When $f'(x) < 0$, the function slopes downward.

Flashcard 31: What is the significance of a sign change in $f'(x)$ at a critical point?

Answer: Indicates a local extremum. Sign changes indicate transitions between increasing/decreasing.

Flashcard 32: What does $f''(x) > 0$ indicate about $f(x)$?

Answer: $f(x)$ is concave up. Positive second derivative means curve bends upward.

Flashcard 33: What is the derivative of $f(x) = x^3 - 3x^2$?

Answer: $f'(x) = 3x^2 - 6x$. Power rule: bring down exponent and reduce by 1.

Flashcard 34: What does $f''(x) > 0$ indicate about $f(x)$?

Answer: $f(x)$ is concave up. Positive second derivative means curve bends upward.

Flashcard 35: What is the second derivative of $f(x) = x^3 - 3x^2 + 2x$?

Answer: $f''(x) = 6x - 6$. Take derivative of $f'(x) = 3x^2 - 6x + 2$.

Flashcard 36: What does $f'(x) = 0$ but no sign change indicate at a critical point?

Answer: No local extremum. No sign change means no local maximum or minimum.

Flashcard 37: What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$?

Answer: $f(x)$ is strictly decreasing on $(a, b)$. Negative derivative throughout means consistently falling.

Flashcard 38: What is the relationship between critical points and intervals of increase/decrease?

Answer: Critical points separate intervals. They divide the domain into regions with consistent behavior.

Flashcard 39: What does $f'(x) = (x-1)^2$ imply about intervals of increase/decrease?

Answer: $f(x)$ is non-decreasing. $f'(x) = (x-1)^2 \geq 0$, never decreasing.

Flashcard 40: What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$?

Answer: $f(x)$ is strictly increasing on $(a, b)$. Positive derivative throughout means consistently rising.

Flashcard 41: Find intervals where $f(x) = x^2 - 8x + 15$ is increasing.

Answer: $x \in (4, \infty)$. $f'(x) = 2x - 8$, positive when $x > 4$.

Flashcard 42: What does $f''(x) < 0$ indicate about $f(x)$?

Answer: $f(x)$ is concave down. Negative second derivative means curve bends downward.

Flashcard 43: What is the test to determine if a function is increasing on an interval?

Answer: The first derivative is positive. When $f'(x) > 0$, the function slopes upward.

Flashcard 44: What does a positive $f'(x)$ indicate about $f(x)$ on an interval?

Answer: $f(x)$ is increasing. Positive slope means function values are rising.

Flashcard 45: What is the significance of a sign change in $f'(x)$ at a critical point?

Answer: Indicates a local extremum. Sign changes indicate transitions between increasing/decreasing.

Flashcard 46: For $f'(x) = 4x^3 - 12x$, find the critical points.

Answer: $x = 0, \pm \sqrt{3}$. $f'(x) = 4x(x^2 - 3)$, set equal to zero.

Flashcard 47: What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$?

Answer: $f(x)$ is strictly decreasing on $(a, b)$. Negative derivative throughout means consistently falling.

Flashcard 48: Identify the intervals of increase for $f(x) = x^3 - 3x^2 + 4$.

Answer: $x \in (2, \infty)$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, positive when $x > 2$.

Flashcard 49: Identify the intervals of increase for $f(x) = x^3 - 3x^2 + 4$.

Answer: $x \in (2, \infty)$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, positive when $x > 2$.

Flashcard 50: What does $f'(x) = 0$ at $x = a$ indicate?

Answer: Possible local extremum. Critical points may be local maxima, minima, or inflection points.

Flashcard 51: What is the condition for $f(x)$ to be constant on an interval?

Answer: $f'(x) = 0$ for all $x$ in that interval. Zero derivative means no rate of change.

Flashcard 52: If $f'(x) < 0$ and $f''(x) > 0$, what can be said about $f(x)$?

Answer: $f(x)$ is decreasing and concave up. Function falls while curving upward like a valley.