3.3: How Derivatives Affect the Shape of a Graph

Goals:

Find out what the first derivative $f'(x)$ tells us about $f(x)$, and
Find out what the second derivative $f''(x)$ tells us about $f(x)$.

What $f'(x)$ tells us about $f(x)$

A function is increasing on an interval $(a, b)$ if for any two numbers $x_1$ and $x_2$ in $(a, b)$, we have $f(x_1) < f(x_2)$ whenever $x_1 < x_2$.
A function is decreasing on an interval $(a, b)$ if for any two numbers $x_1, x_2 \in (a, b)$, we have $f(x_1) > f(x_2)$ whenever $x_1 < x_2$.

The derivative tells us exactly where a function is increasing and decreasing.

Increasing/Decreasing Test

If $f'(x) > 0$ for $x \in (a, b)$ then $f$ is increasing on $(a, b)$.
If $f'(x) < 0$ for $x \in (a, b)$ then $f$ is decreasing on $(a, b)$.

To use the I/D Test, first find critical numbers, then check signs of $f'(x)$ between critical numbers. This works since the derivative may change sign only when you move past a critical number.

Find intervals where $f(x) = x^2$ is increasing.

Determine the intervals where \[f(x) = x + \dfrac{1}{x}\] is increasing/decreasing.

Find where the function $f(x) = 3x^4 - 4x^3 - 12x^2 + 5$ is increasing or decreasing.

In the previous section we were concerned with finding absolute extrema. How about local extrema?

The First Derivative Test Suppose $c$ is a critical number of a continuous function $f$.

If $f'$ changes from positive to negative at $c$, then $f$ has a local maximum at $c$.
If $f'$ changes from negative to positive at $c$, then $f$ has a local minimum at $c$.
If $f'$ changes doesn't change sign at $c$, then $f$ has no local maximum or minimum at $c$.

Using the First Derivative Test is the same method as the I/D Test. Just make sure the function is continuous.

Find the local minimum and local maximum values for the previous function.

Find the local minimum and local maximum values for the function $f(x) = x^4 - 4x^3$.

What $f''(x)$ tells us about $f(x)$

The second derivative also tells us more information about a function.

A graph is concave up on an interval $I$ if all of the tangent lines lie below the graph on $I$. A graph is concave down on an interval $I$ if all of the tangents lie above the graph on $I$.

An inflection point is a point $P$ on the graph of $f(x)$ where

$f(x)$ is continuous at that point and
the curve changes from concave up to concave down or vice versa.

Concavity Test

If $f''(x) > 0$ for all $x$ in an interval $I$, then the graph of $f$ is concave upward on $I$.
If $f''(x) < 0$ for all $x$ in an interval $I$, then the graph of $f$ is concave downward on $I$.

To use the Concavity Test, we employ a similar method as I/D Test:

Solve $f''(x) = 0$ and find where $f''(x)$ does not exist. Save these numbers.
Create a sign diagram of the second derivative $f''(x)$ by putting numbers from step 1 on a real number line, creating intervals. Then plug in numbers from each interval into $f\ ''(x)$.

Find the intervals of concavity for the function $f(x) = x^4 - 4x^3$.

Warning: It is not necessary for all solutions of $f''(x) = 0$ to be inflection points.

This is similar to the critical number case where $f'(x)$ DNE; you may have not found a local maximum or minimum and found a vertical tangent instead.

Find the intervals of concavity for the function $f(x) = x^4 - x$.

Find the intervals of concavity for the function \[g(x) = \tan x, \quad x \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\]

Sketch a possible graph of a function which satisfies the following:

$f(0) = 0, \ f(2) = 3, \ f(4) = 6, \ f'(0) = f'(4) = 0$
$f'(x) > 0$ for $0 < x < 4 \ , f'(x) < 0$ for $x < 0$ and $x > 4$
$f''(x) > 0$ for $x < 2 \ , f''(x) < 0$ for $ x > 2$.

$f''(x)$ can also be used to find local minimums and maximums. The framework is similar to the First Derivative Test:

Find potential locations of local extrema (critical numbers).
Use $f''(x)$ to determine if the location contains a local minimum or maximum.

Second Derivative Test
Suppose $f''(x)$ is continuous near $c$.

If $f'(c) = 0$ and $f''(c) > 0$, then $f$ has a local minimum at $c$.
If $f'(c) = 0$ and $f''(c) < 0$, then $f$ has a local maximum at $c$.
If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive. Test the critical number with the First Derivative Test instead.

Find all local minimums and maximums for the function $f(x) = x^4 - 4x^3$ using the Second Derivative Test.

Find all local minimums and maximums for the function $f(x) = -x^4 + 2x^2 + 2$ using the Second Derivative Test.