Topics Covered:
Critical Points: Points where \(f'(c) = 0\) or \(f'(c)\) does not exist (DNE).
Absolute min/max: Can occur at endpoints or critical points.
Relative/Local min/max: Only occur at critical points.
First Derivative Test:
For a Relative Maximum at \(x = c\):
For a Relative Minimum at \(x = c\):
Graph Description: A continuous curve showing multiple critical points. At local maxima, the curve rises from the left and falls to the right, with horizontal tangent lines at the peaks. At local minima, the curve falls from the left and rises to the right, with horizontal tangent lines at the valleys.
Step 1: Find the derivative
Using the product rule:
\[f'(x) = e^{-x} + x \cdot \frac{d}{dx}[e^{-x}] = e^{-x} + x \cdot (-e^{-x}) = e^{-x}(1 - x)\]Step 2: Find critical points by setting \(f'(x) = 0\)
\[e^{-x}(1 - x) = 0\]Since \(e^{-x} \neq 0\) for all \(x\), we have:
\[1 - x = 0 \implies x = 1\]So \(x = 1\) is the only critical point, with \(f'(1) = 0\).
Step 3: Create a sign chart for \(f'(x) = e^{-x}(1-x)\)
| Interval | Sign of \((1-x)\) | Sign of \(f'(x)\) | Behavior |
|---|---|---|---|
| \((-\infty, 1)\) | \(+\) | \(+\) | Increasing ↑ |
| \((1, \infty)\) | \(-\) | \(-\) | Decreasing ↓ |
Conclusion: Since \(f'(x)\) changes from positive to negative at \(x = 1\), there is a local maximum at \(x = 1\).
The second derivative \(f''(x) = \frac{d}{dx}[f'(x)]\) provides information about how the first derivative is changing.
Relationship Between \(f''(x)\) and \(f'(x)\):
Concave Up: A function is concave up on an interval where \(f''(x) > 0\). The graph curves upward, like a cup (U shape). The slope \(f'(x)\) is increasing.
Concave Down: A function is concave down on an interval where \(f''(x) < 0\). The graph curves downward, like a cap (∩ shape). The slope \(f'(x)\) is decreasing.
Visual Examples:
The Second Derivative Test provides an alternative method to determine whether a critical point is a local maximum or minimum, without needing to check the sign of \(f'(x)\) on both sides.
Second Derivative Test:
For a Relative/Local Maximum at \(x = c\):
For a Relative/Local Minimum at \(x = c\):
Intuition: At a local maximum, the function is concave down (looks like ∩), so \(f''(c) < 0\). At a local minimum, the function is concave up (looks like U), so \(f''(c) > 0\).
What happens when \(f''(c) = 0\)?
When the second derivative equals zero at a critical point, the Second Derivative Test is inconclusive. You must use the First Derivative Test instead.
Case 1: \(f(x) = x^4\)
Case 2: \(f(x) = -x^4\)
Case 3: \(f(x) = x^3\)
Point of Inflection: A point where the concavity of the function changes. If \(f(x)\) is continuous, \(f'(x)\) is continuous and differentiable, then points where \(f''(x) = 0\) are called points of inflection.
At a point of inflection, the function changes from concave up to concave down, or vice versa.
Important Note: Not every point where \(f''(x) = 0\) is necessarily a point of inflection. You must verify that the concavity actually changes on either side of the point.
Step 1: Find the first derivative (from earlier)
\[f'(x) = e^{-x}(1-x)\]Step 2: Find the second derivative
Using the product rule on \(f'(x) = e^{-x}(1-x)\):
\[f''(x) = -e^{-x}(1-x) + e^{-x}(-1)\] \[f''(x) = -e^{-x} + xe^{-x} - e^{-x}\] \[f''(x) = e^{-x}(x - 2)\]Step 3: Find potential inflection points by setting \(f''(x) = 0\)
\[e^{-x}(x - 2) = 0\]Since \(e^{-x} \neq 0\), we have:
\[x - 2 = 0 \implies x = 2\]So \(x = 2\) is the only potential inflection point.
Step 4: Determine concavity on each interval
| Interval | Test Point | Sign of \((x-2)\) | Sign of \(f''(x)\) | Concavity |
|---|---|---|---|---|
| \((-\infty, 2)\) | \(x = 0\) | \(-\) | \(-\) | Concave Down |
| \((2, \infty)\) | \(x = 3\) | \(+\) | \(+\) | Concave Up |
Conclusion:
We found that \(x = 1\) is a critical point with \(f'(1) = 0\).
Apply the Second Derivative Test:
We have \(f''(x) = e^{-x}(x-2)\), so:
\[f''(1) = e^{-1}(1-2) = e^{-1}(-1) = -\frac{1}{e} < 0\]Since \(f''(1) < 0\), the function is concave down at \(x = 1\).
Conclusion: \(x = 1\) is a local maximum. ✓
This confirms our result from the First Derivative Test!
Complete Graph Behavior: The function \(f(x) = xe^{-x}\) increases on \((-\infty, 1)\), has a local maximum at \(x = 1\), decreases on \((1, \infty)\), is concave down on \((-\infty, 2)\), has an inflection point at \(x = 2\), and is concave up on \((2, \infty)\).
Step 1: Find critical points
\[f'(x) = 12x^3 - 12x^2 - 24x = 0\] \[12x(x^2 - x - 2) = 0\] \[12x(x + 1)(x - 2) = 0\] \[x = -1, \quad x = 0, \quad x = 2\]We have 3 critical points.
Step 2: Find the second derivative
\[f''(x) = 36x^2 - 24x - 24\]Step 3: Apply the Second Derivative Test at each critical point
At \(x = -1\):
\[f''(-1) = 36(-1)^2 - 24(-1) - 24 = 36 + 24 - 24 = 36 > 0\]Since \(f''(-1) > 0\) (concave up), \(x = -1\) is a local minimum. ✓
At \(x = 0\):
\[f''(0) = 36(0)^2 - 24(0) - 24 = -24 < 0\]Since \(f''(0) < 0\) (concave down), \(x = 0\) is a local maximum. ✓
At \(x = 2\):
\[f''(2) = 36(2)^2 - 24(2) - 24 = 36(4) - 48 - 24 = 144 - 48 - 24 = 72 > 0\]Since \(f''(2) > 0\) (concave up), \(x = 2\) is a local minimum. ✓
Key Relationships:
| Condition | Meaning |
|---|---|
| \(f'(x) > 0\) | Function is increasing |
| \(f'(x) < 0\) | Function is decreasing |
| \(f'(x) = 0\) | Horizontal tangent (potential max/min) |
| \(f''(x) > 0\) | Concave up (slopes increasing) |
| \(f''(x) < 0\) | Concave down (slopes decreasing) |
| \(f''(x) = 0\) | Potential inflection point |
Two Methods for Finding Local Extrema: