Connecting a Function
and Its Derivatives
Discover how \( f \), \( f' \), and \( f'' \) work together to reveal a function's increasing/decreasing behavior, concavity, and key features.
AP Calculus BC
What Does \( f'(x) \) Tell Us?
The first derivative \( f'(x) \) measures the instantaneous rate of change of \( f(x) \). Its sign directly tells us about the original function's behavior:
\( f'(x) > 0 \)
\( f \) is increasing on that interval
\( f'(x) < 0 \)
\( f \) is decreasing on that interval
\( f'(x) = 0 \)
Possible local max, local min, or neither
What Does \( f''(x) \) Tell Us?
The second derivative \( f''(x) \) measures the rate of change of the slope. It reveals the concavity of the original function:
\( f''(x) > 0 \)
\( f \) is concave up (slopes are increasing)
\( f''(x) < 0 \)
\( f \) is concave down (slopes are decreasing)
\( f''(x) = 0 \)
Possible inflection point (concavity may change)
The Big Picture: Three Functions, One Story
The First Derivative Test
If \( f'(c) = 0 \) or \( f'(c) \) is undefined, then \( x = c \) is a critical point. To classify it:
1
If \( f' \) changes from positive to negative at \( c \), then \( f(c) \) is a local maximum.
2
If \( f' \) changes from negative to positive at \( c \), then \( f(c) \) is a local minimum.
3
If \( f' \) does not change sign at \( c \), then \( f(c) \) is neither a max nor a min.
The Second Derivative Test
An alternative way to classify critical points using concavity:
1
Find \( c \) where \( f'(c) = 0 \).
2
Evaluate \( f''(c) \).
3
If \( f''(c) > 0 \): concave up → local minimum.
4
If \( f''(c) < 0 \): concave down → local maximum.
5
If \( f''(c) = 0 \): test is inconclusive; use the First Derivative Test.
First vs Second Derivative Test
| First Derivative Test | Second Derivative Test | |
|---|---|---|
| Requires | Sign chart of \( f' \) | Value of \( f''(c) \) |
| Always works? | Yes | No (inconclusive if \( f''(c) = 0 \)) |
| Speed | Requires interval analysis | Often faster: one computation |
| Best for | Non-differentiable critical points | Polynomial / smooth functions |
Finding Inflection Points
An inflection point is where \( f \) changes concavity. To find them:
1
Find where \( f''(x) = 0 \) or \( f''(x) \) is undefined.
2
Check that \( f'' \) changes sign across that point.
3
If yes: inflection point. If no: not an inflection point.
Step-by-Step Analysis Strategy
1
Find \( f'(x) \) and set \( f'(x) = 0 \) to locate critical points.
2
Build a sign chart for \( f' \) to determine increasing/decreasing intervals.
3
Classify each critical point using the First or Second Derivative Test.
4
Find \( f''(x) \) and set \( f''(x) = 0 \) to locate candidate inflection points.
5
Build a sign chart for \( f'' \) to determine concavity intervals and confirm inflection points.
6
Combine all information to sketch or describe the full behavior of \( f \).
Example 1: \( f(x) = x^3 - 3x \)
Graph of \( f(x) = x^3 - 3x \)
Example 2: \( g(x) = xe^{-x} \)
Example 3: Reading the Graph of \( f' \)
Common Mistakes to Avoid
✕ Assuming \( f''(c) = 0 \) means inflection
\( f''(c) = 0 \) is necessary but not sufficient. You must verify that \( f'' \) changes sign at \( c \).
✕ Confusing \( f \) with \( f' \)
When \( f' > 0 \), it means \( f \) is increasing, not that \( f \) is positive. The sign of \( f' \) describes slope, not value.
✕ Forgetting endpoints
On a closed interval, absolute extrema may occur at endpoints. The derivative tests only find local extrema.
Summary Cheat Sheet
| Condition | \( f \) behavior | \( f' \) sign | \( f'' \) sign |
|---|---|---|---|
| Increasing, concave up | Rising, curving upward | \(+\) | \(+\) |
| Increasing, concave down | Rising, curving downward | \(+\) | \(-\) |
| Decreasing, concave up | Falling, curving upward | \(-\) | \(+\) |
| Decreasing, concave down | Falling, curving downward | \(-\) | \(-\) |
| Local maximum | Peak | \(0\) (\(+\) to \(-\)) | \(-\) |
| Local minimum | Valley | \(0\) (\(-\) to \(+\)) | \(+\) |
| Inflection point | Concavity switch | Any | \(0\) (sign change) |
Question 1 of 5
If \( f'(x) > 0 \) and \( f''(x) < 0 \) on an interval, what is \( f \) doing?
\( f' > 0 \) means increasing. \( f'' < 0 \) means concave down. Think of a hill that is still rising but starting to flatten out.
Question 2 of 5
Let \( f(x) = x^4 - 4x^3 \). At what value of \( x \) does \( f \) have a local minimum?
\( f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) \). Critical points at \( x = 0 \) and \( x = 3 \). At \( x = 3 \): \( f' \) changes from negative to positive, so it is a local minimum. At \( x = 0 \), \( f' \) does not change sign (stays non-positive), so no extremum there.
Question 3 of 5
If the graph of \( f' \) crosses the \( x \)-axis from positive to negative at \( x = c \), what happens to \( f \) at \( x = c \)?
By the First Derivative Test: when \( f' \) changes from positive to negative, \( f \) transitions from increasing to decreasing, creating a local maximum.
Question 4 of 5
For \( h(x) = x^3 \), what can you say about the point \( x = 0 \)?
\( h'(x) = 3x^2 \), so \( h'(0) = 0 \). But \( h' \ge 0 \) everywhere: it does not change sign. So \( x = 0 \) is not an extremum. However, \( h''(x) = 6x \) changes from negative to positive at \( x = 0 \), making it an inflection point.
Question 5 of 5
The Second Derivative Test is inconclusive when:
The Second Derivative Test requires \( f'(c) = 0 \) and then examines \( f''(c) \). When both \( f'(c) = 0 \) and \( f''(c) = 0 \), the test cannot determine the nature of the critical point: you must use the First Derivative Test instead.