Determining Intervals on Increasing, Decreasing Functions - AP Calculus BC
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What is the relationship between critical points and intervals of increase/decrease?
What is the relationship between critical points and intervals of increase/decrease?
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Critical points separate intervals. They divide the domain into regions with consistent behavior.
Critical points separate intervals. They divide the domain into regions with consistent behavior.
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What is the test to determine if a function is increasing on an interval?
What is the test to determine if a function is increasing on an interval?
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The first derivative is positive. When $f'(x) > 0$, the function slopes upward.
The first derivative is positive. When $f'(x) > 0$, the function slopes upward.
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How can you determine the critical points of $f(x) = x^3 - 6x^2$?
How can you determine the critical points of $f(x) = x^3 - 6x^2$?
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Solve $f'(x) = 0$: $x = 0, 4$. $f'(x) = 3x^2 - 12x = 3x(x-4)$, set equal to zero.
Solve $f'(x) = 0$: $x = 0, 4$. $f'(x) = 3x^2 - 12x = 3x(x-4)$, set equal to zero.
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What is the condition for $f(x)$ to be constant on an interval?
What is the condition for $f(x)$ to be constant on an interval?
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$f'(x) = 0$ for all $x$ in that interval. Zero derivative means no rate of change.
$f'(x) = 0$ for all $x$ in that interval. Zero derivative means no rate of change.
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What does a negative $f'(x)$ indicate about $f(x)$ on an interval?
What does a negative $f'(x)$ indicate about $f(x)$ on an interval?
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$f(x)$ is decreasing. Negative slope means function values are falling.
$f(x)$ is decreasing. Negative slope means function values are falling.
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For $f'(x) = 4x^3 - 12x$, find the critical points.
For $f'(x) = 4x^3 - 12x$, find the critical points.
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$x = 0, \pm \sqrt{3}$. $f'(x) = 4x(x^2 - 3)$, set equal to zero.
$x = 0, \pm \sqrt{3}$. $f'(x) = 4x(x^2 - 3)$, set equal to zero.
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What does $f''(x) < 0$ indicate about $f(x)$?
What does $f''(x) < 0$ indicate about $f(x)$?
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$f(x)$ is concave down. Negative second derivative means curve bends downward.
$f(x)$ is concave down. Negative second derivative means curve bends downward.
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State the critical point condition for a function $f(x)$.
State the critical point condition for a function $f(x)$.
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$f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.
$f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.
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Find intervals where $f(x) = x^2 - 8x + 15$ is decreasing.
Find intervals where $f(x) = x^2 - 8x + 15$ is decreasing.
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$x \in (-\infty, 4)$. $f'(x) = 2x - 8$, negative when $x < 4$.
$x \in (-\infty, 4)$. $f'(x) = 2x - 8$, negative when $x < 4$.
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If $f'(x) < 0$ and $f''(x) > 0$, what can be said about $f(x)$?
If $f'(x) < 0$ and $f''(x) > 0$, what can be said about $f(x)$?
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$f(x)$ is decreasing and concave up. Function falls while curving upward like a valley.
$f(x)$ is decreasing and concave up. Function falls while curving upward like a valley.
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What does $f'(x) = (x-1)^2$ imply about intervals of increase/decrease?
What does $f'(x) = (x-1)^2$ imply about intervals of increase/decrease?
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$f(x)$ is non-decreasing. $f'(x) = (x-1)^2 \geq 0$, never decreasing.
$f(x)$ is non-decreasing. $f'(x) = (x-1)^2 \geq 0$, never decreasing.
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What does a positive $f'(x)$ indicate about $f(x)$ on an interval?
What does a positive $f'(x)$ indicate about $f(x)$ on an interval?
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$f(x)$ is increasing. Positive slope means function values are rising.
$f(x)$ is increasing. Positive slope means function values are rising.
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For $f'(x) = -4x$, on which interval is $f(x)$ decreasing?
For $f'(x) = -4x$, on which interval is $f(x)$ decreasing?
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$x \in (0, \infty)$. $f'(x) = -4x < 0$ when $x > 0$.
$x \in (0, \infty)$. $f'(x) = -4x < 0$ when $x > 0$.
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Find intervals where $f(x) = x^2 - 8x + 15$ is increasing.
Find intervals where $f(x) = x^2 - 8x + 15$ is increasing.
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$x \in (4, \infty)$. $f'(x) = 2x - 8$, positive when $x > 4$.
$x \in (4, \infty)$. $f'(x) = 2x - 8$, positive when $x > 4$.
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What is the second derivative of $f(x) = x^3 - 3x^2 + 2x$?
What is the second derivative of $f(x) = x^3 - 3x^2 + 2x$?
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$f''(x) = 6x - 6$. Take derivative of $f'(x) = 3x^2 - 6x + 2$.
$f''(x) = 6x - 6$. Take derivative of $f'(x) = 3x^2 - 6x + 2$.
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What does $f'(x) = 0$ but no sign change indicate at a critical point?
What does $f'(x) = 0$ but no sign change indicate at a critical point?
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No local extremum. No sign change means no local maximum or minimum.
No local extremum. No sign change means no local maximum or minimum.
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What does $f''(x) > 0$ indicate about $f(x)$?
What does $f''(x) > 0$ indicate about $f(x)$?
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$f(x)$ is concave up. Positive second derivative means curve bends upward.
$f(x)$ is concave up. Positive second derivative means curve bends upward.
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What does $f'(x) = 0$ at $x = a$ indicate?
What does $f'(x) = 0$ at $x = a$ indicate?
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Possible local extremum. Critical points may be local maxima, minima, or inflection points.
Possible local extremum. Critical points may be local maxima, minima, or inflection points.
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What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$?
What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$?
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$f(x)$ is strictly increasing on $(a, b)$. Positive derivative throughout means consistently rising.
$f(x)$ is strictly increasing on $(a, b)$. Positive derivative throughout means consistently rising.
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What is the significance of a sign change in $f'(x)$ at a critical point?
What is the significance of a sign change in $f'(x)$ at a critical point?
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Indicates a local extremum. Sign changes indicate transitions between increasing/decreasing.
Indicates a local extremum. Sign changes indicate transitions between increasing/decreasing.
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What is a necessary condition for a point to be a local extremum?
What is a necessary condition for a point to be a local extremum?
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$f'(x) = 0$ or $f'(x)$ is undefined. Local extrema occur only at critical points.
$f'(x) = 0$ or $f'(x)$ is undefined. Local extrema occur only at critical points.
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Identify the intervals of increase for $f(x) = x^3 - 3x^2 + 4$.
Identify the intervals of increase for $f(x) = x^3 - 3x^2 + 4$.
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$x \in (2, \infty)$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, positive when $x > 2$.
$x \in (2, \infty)$. $f'(x) = 3x^2 - 6x = 3x(x-2)$, positive when $x > 2$.
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What is the test to determine if a function is decreasing on an interval?
What is the test to determine if a function is decreasing on an interval?
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The first derivative is negative. When $f'(x) < 0$, the function slopes downward.
The first derivative is negative. When $f'(x) < 0$, the function slopes downward.
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What is the derivative of $f(x) = x^3 - 3x^2$?
What is the derivative of $f(x) = x^3 - 3x^2$?
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$f'(x) = 3x^2 - 6x$. Power rule: bring down exponent and reduce by 1.
$f'(x) = 3x^2 - 6x$. Power rule: bring down exponent and reduce by 1.
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What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$?
What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$?
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$f(x)$ is strictly decreasing on $(a, b)$. Negative derivative throughout means consistently falling.
$f(x)$ is strictly decreasing on $(a, b)$. Negative derivative throughout means consistently falling.
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For $f'(x) = 4x^3 - 12x$, find the critical points.
For $f'(x) = 4x^3 - 12x$, find the critical points.
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$x = 0, \pm \sqrt{3}$. $f'(x) = 4x(x^2 - 3)$, set equal to zero.
$x = 0, \pm \sqrt{3}$. $f'(x) = 4x(x^2 - 3)$, set equal to zero.
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What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$?
What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$?
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$f(x)$ is strictly increasing on $(a, b)$. Positive derivative throughout means consistently rising.
$f(x)$ is strictly increasing on $(a, b)$. Positive derivative throughout means consistently rising.
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State the critical point condition for a function $f(x)$.
State the critical point condition for a function $f(x)$.
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$f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.
$f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.
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What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$?
What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$?
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$f(x)$ is strictly decreasing on $(a, b)$. Negative derivative throughout means consistently falling.
$f(x)$ is strictly decreasing on $(a, b)$. Negative derivative throughout means consistently falling.
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What is the derivative of $f(x) = x^3 - 3x^2$?
What is the derivative of $f(x) = x^3 - 3x^2$?
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$f'(x) = 3x^2 - 6x$. Power rule: bring down exponent and reduce by 1.
$f'(x) = 3x^2 - 6x$. Power rule: bring down exponent and reduce by 1.
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