First Derivative Test
Relative Extrema
Master the technique of using sign changes in \( f'(x) \) to classify local maxima and local minima of a function.
AP Calculus BC
What Are Relative Extrema?
A function \( f \) has a relative (local) maximum at \( x = c \) if \( f(c) \geq f(x) \) for all \( x \) near \( c \). Similarly, it has a relative minimum at \( x = c \) if \( f(c) \leq f(x) \) for all \( x \) near \( c \).
Local Maximum
The function value is greater than or equal to nearby values: a "hilltop."
Local Minimum
The function value is less than or equal to nearby values: a "valley."
Visual Intuition
The Role of Critical Numbers
A number \( c \) in the domain of \( f \) is a critical number if either \( f'(c) = 0 \) or \( f'(c) \) does not exist. Relative extrema can only occur at critical numbers.
\( f'(c) = 0 \)
The tangent line is horizontal: a smooth turning point.
\( f'(c) \) DNE
A cusp, corner, or vertical tangent where the derivative is undefined.
The First Derivative Test
Let \( c \) be a critical number of a continuous function \( f \).
1
If \( f' \) changes from positive to negative at \( c \), then \( f(c) \) is a relative maximum.
2
If \( f' \) changes from negative to positive at \( c \), then \( f(c) \) is a relative minimum.
3
If \( f' \) does not change sign at \( c \), then \( f(c) \) is neither a max nor a min.
Reading a Sign Chart
First vs Second Derivative Test
| First Derivative Test | Second Derivative Test | |
|---|---|---|
| What you check | Sign change of \( f' \) | Sign of \( f''(c) \) |
| Requires | \( f' \) exists near \( c \) | \( f''(c) \) exists |
| Inconclusive when | Never (always decisive) | \( f''(c) = 0 \) |
| Also reveals | Intervals of inc/dec | Concavity at \( c \) |
How to Apply the Test
1
Find \( f'(x) \). Differentiate the given function.
2
Find critical numbers. Solve \( f'(x) = 0 \) and identify where \( f'(x) \) is undefined (but \( f \) is defined).
3
Build a sign chart. Choose test values in each interval determined by the critical numbers. Evaluate the sign of \( f' \).
4
Classify. \( + \rightarrow - \) means relative max. \( - \rightarrow + \) means relative min. No sign change means neither.
5
State the value. Compute \( f(c) \) for each extremum to give the coordinate.
Example 1: \( f(x) = x^3 - 3x + 2 \)
1
Differentiate: \( f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1) \)
2
Critical numbers: \( f'(x) = 0 \) when \( x = -1 \) and \( x = 1 \)
3
Sign chart: Test \( x = -2 \): \( f'(-2) = 9 > 0 \). Test \( x = 0 \): \( f'(0) = -3 < 0 \). Test \( x = 2 \): \( f'(2) = 9 > 0 \).
Graph of \( f(x) = x^3 - 3x + 2 \)
Example 2: \( g(x) = x^{2/3}(x - 4) \)
1
Expand: \( g(x) = x^{5/3} - 4x^{2/3} \)
2
Differentiate: \( g'(x) = \frac{5}{3}x^{2/3} - \frac{8}{3}x^{-1/3} = \frac{5x - 8}{3x^{1/3}} \)
3
Critical numbers: \( g'(x) = 0 \) when \( x = \frac{8}{5} \). Also \( g'(0) \) is undefined (but \( g(0) = 0 \) is defined), so \( x = 0 \) is critical too.
Example 3: \( h(x) = \sin x + \cos x \)
Find the relative extrema on \( (0, 2\pi) \).
1
Differentiate: \( h'(x) = \cos x - \sin x \)
2
Solve \( h'(x) = 0 \): \( \cos x = \sin x \implies x = \frac{\pi}{4},\; \frac{5\pi}{4} \)
3
Sign chart: Test \( x \approx 0.1 \): \( h'(x) > 0 \). Test \( x = 2 \): \( h'(x) < 0 \). Test \( x = 5 \): \( h'(x) > 0 \).
Why Sign Changes Matter
Special Cases to Know
Cusp at \( c \)
\( f'(c) \) DNE, but if the sign of \( f' \) changes across \( c \), the First Derivative Test still applies.
Endpoint Extrema
On a closed interval \( [a,b] \), check endpoints separately. The FDT is for interior critical points only.
Piecewise Functions
Where pieces join, \( f' \) may not exist. Treat those join points as critical numbers and test signs on each side.
Common Mistakes
Assuming \( f'(c) = 0 \) means extremum
Not all critical numbers yield extrema. You must verify a sign change in \( f' \).
Forgetting where \( f' \) is undefined
Critical numbers include points where \( f' \) DNE but \( f \) is defined. Don't just solve \( f'(x) = 0 \).
Testing values outside the domain
If the domain is restricted (e.g., \( x > 0 \)), only pick test values within the domain of \( f \).
Quick Reference Summary
| Sign of \( f' \) before \( c \) | Sign of \( f' \) after \( c \) | Conclusion |
|---|---|---|
| \(+\) | \(-\) | Relative Maximum |
| \(-\) | \(+\) | Relative Minimum |
| \(+\) | \(+\) | Neither (no extremum) |
| \(-\) | \(-\) | Neither (no extremum) |
Question 1 of 5
If \( f'(x) \) changes from negative to positive at \( x = c \), then \( f(c) \) is a:
When \( f' \) changes from negative to positive, \( f \) switches from decreasing to increasing, creating a valley: a relative minimum.
Question 2 of 5
Let \( f(x) = x^4 - 4x^3 \). At which critical number does \( f \) have a relative minimum?
\( f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3) \). Critical numbers: \( x = 0 \) and \( x = 3 \). At \( x = 3 \), \( f' \) changes from \( - \) to \( + \), giving a relative min. At \( x = 0 \), the sign does not change (\( - \) to \( - \)): neither.
Question 3 of 5
\( f'(x) = (x - 2)^2(x + 1) \). What happens at \( x = 2 \)?
Since \( (x - 2)^2 \geq 0 \) always, the sign of \( f'(x) \) near \( x = 2 \) depends only on \( (x + 1) \). For \( x \) near 2, \( (x + 1) > 0 \), so \( f' > 0 \) on both sides. No sign change → neither max nor min.
Question 4 of 5
Which test always gives a conclusive answer at a critical number where \( f \) is continuous?
The First Derivative Test examines the sign change of \( f' \) and always classifies the critical number as a max, min, or neither. The Second Derivative Test is inconclusive when \( f''(c) = 0 \).
Question 5 of 5
For \( h(x) = \sin x + \cos x \) on \( (0, 2\pi) \), the relative maximum value is:
\( h'(x) = \cos x - \sin x = 0 \) at \( x = \frac{\pi}{4} \). Since \( f' \) goes from \( + \) to \( - \), this is a relative max. The value is \( h\!\left(\frac{\pi}{4}\right) = \sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \).