Determining Intervals on Increasing, Decreasing Functions - AP Calculus BC

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

Solve $f'(x) = 0$ : $x = 0, 4$ . $f'(x) = 3x^2 - 12x = 3x(x-4)$ , set equal to zero.
Find intervals where $f(x) = x^2 - 8x + 15$ is increasing.

$x \in (4, \infty)$ . $f'(x) = 2x - 8$ , positive when $x > 4$ .
What is the second derivative of $f(x) = x^3 - 3x^2 + 2x$ ?

$f''(x) = 6x - 6$ . Take derivative of $f'(x) = 3x^2 - 6x + 2$ .
If $f'(x) < 0$ and $f''(x) > 0$ , what can be said about $f(x)$ ?

$f(x)$ is decreasing and concave up. Function falls while curving upward like a valley.
For $f'(x) = -4x$ , on which interval is $f(x)$ decreasing?

$x \in (0, \infty)$ . $f'(x) = -4x < 0$ when $x > 0$ .
What does $f'(x) = (x-1)^2$ imply about intervals of increase/decrease?

$f(x)$ is non-decreasing. $f'(x) = (x-1)^2 \geq 0$ , never decreasing.
State the critical point condition for a function $f(x)$ .

$f'(x) = 0$ or $f'(x)$ is undefined. Critical points occur where slope is zero or doesn't exist.
Find intervals where $f(x) = x^2 - 8x + 15$ is decreasing.

$x \in (-\infty, 4)$ . $f'(x) = 2x - 8$ , negative when $x < 4$ .
What does a negative $f'(x)$ indicate about $f(x)$ on an interval?

$f(x)$ is decreasing. Negative slope means function values are falling.
What is the test to determine if a function is decreasing on an interval?

The first derivative is negative. When $f'(x) < 0$ , the function slopes downward.
What does $f'(x) = 0$ at $x = a$ indicate?

Possible local extremum. Critical points may be local maxima, minima, or inflection points.
What does a positive $f'(x)$ indicate about $f(x)$ on an interval?

$f(x)$ is increasing. Positive slope means function values are rising.
What is the first derivative test used for?

To determine increasing or decreasing intervals. Analyzes sign of $f'(x)$ to find where function rises or falls.
What is the relationship between critical points and intervals of increase/decrease?

Critical points separate intervals. They divide the domain into regions with consistent behavior.
For $f'(x) = 4x^3 - 12x$ , find the critical points.

$x = 0, \pm \sqrt{3}$ . $f'(x) = 4x(x^2 - 3)$ , set equal to zero.
What is the derivative of $f(x) = x^3 - 3x^2$ ?

$f'(x) = 3x^2 - 6x$ . Power rule: bring down exponent and reduce by 1.
What is a necessary condition for a point to be a local extremum?

$f'(x) = 0$ or $f'(x)$ is undefined. Local extrema occur only at critical points.
What is the second derivative test used for?

To determine concavity. Analyzes $f''(x)$ to determine curve bending direction.
What does $f''(x) < 0$ indicate about $f(x)$ ?

$f(x)$ is concave down. Negative second derivative means curve bends downward.
What does $f'(x) = 0$ but no sign change indicate at a critical point?

No local extremum. No sign change means no local maximum or minimum.
What does $f''(x) > 0$ indicate about $f(x)$ ?

$f(x)$ is concave up. Positive second derivative means curve bends upward.
What does it mean if $f'(x) < 0$ for all $x$ in $(a, b)$ ?

$f(x)$ is strictly decreasing on $(a, b)$ . Negative derivative throughout means consistently falling.
What does it mean if $f'(x) > 0$ for all $x$ in $(a, b)$ ?

$f(x)$ is strictly increasing on $(a, b)$ . Positive derivative throughout means consistently rising.
What is the test to determine if a function is increasing on an interval?

The first derivative is positive. When $f'(x) > 0$ , the function slopes upward.
What is the significance of a sign change in $f'(x)$ at a critical point?

Indicates a local extremum. Sign changes indicate transitions between increasing/decreasing.
Identify the intervals of increase for $f(x) = x^3 - 3x^2 + 4$ .

$x \in (2, \infty)$ . $f'(x) = 3x^2 - 6x = 3x(x-2)$ , positive when $x > 2$ .
What is the condition for $f(x)$ to be constant on an interval?

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

Study these flashcards