Calculus I - Week 8
Curve Sketching and the Mean Value Theorem
Derivatives tell us everything about the shape of a curve. This week uses the first and second derivatives to determine where functions increase, decrease, and curve — leading to a complete portrait of any differentiable function. The Mean Value Theorem anchors the analysis with a powerful geometric guarantee.
Learning Objectives
- State and apply the Mean Value Theorem. - Use the first derivative to find intervals of increase and decrease. - Classify critical points using the first and second derivative tests. - Use the second derivative to determine concavity and inflection points. - Identify local maxima and minima from the sign chart. - Produce a complete sketch of a function using derivative information.
1. The Mean Value Theorem
If f is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) such that: f'(c) = (f(b) - f(a)) / (b - a) Geometrically: there is a point where the tangent line is parallel to the secant line from a to b. Interpretation: if you drive from A to B averaging 60 mph, then at some moment you were traveling exactly 60 mph.
2. Increasing and Decreasing
If f'(x) > 0 on an interval, f is increasing there. If f'(x) < 0 on an interval, f is decreasing there. Procedure: 1. Find critical points: solve f'(x) = 0 or find where f'(x) is undefined. 2. Make a sign chart for f'(x). 3. Read off intervals of increase and decrease.
3. First Derivative Test Example
Analyze f(x) = x^3 - 3x. f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1). Critical points: x = -1 and x = 1. Sign chart: - x < -1: f'(+) > 0, increasing - -1 < x < 1: f'(-) < 0, decreasing - x > 1: f'(+) > 0, increasing Local max at x = -1: f(-1) = 2. Local min at x = 1: f(1) = -2.
4. Concavity and the Second Derivative
If f''(x) > 0 on an interval, f is concave up (cup shape). If f''(x) < 0 on an interval, f is concave down (cap shape). An inflection point is where concavity changes — where f'' changes sign. For f(x) = x^3 - 3x: f''(x) = 6x. f''(x) = 0 at x = 0 — inflection point. x < 0: concave down. x > 0: concave up.
5. Second Derivative Test
At a critical point c where f'(c) = 0: - If f''(c) > 0: local minimum (concave up at c). - If f''(c) < 0: local maximum (concave down at c). - If f''(c) = 0: test inconclusive; use the first derivative test. For x = -1: f''(-1) = -6 < 0 → local max. ✓ For x = 1: f''(1) = 6 > 0 → local min. ✓
6. Full Curve Sketching Procedure
1. Find domain. 2. Find intercepts (set y=0 for x-intercepts, x=0 for y-intercept). 3. Find critical points and classify (first or second derivative test). 4. Determine intervals of increase/decrease. 5. Find inflection points from f''=0 with sign change. 6. Determine concavity on each interval. 7. Identify any asymptotes. 8. Plot the key points and sketch.
7. Example: Full Sketch of a Cubic
Sketch f(x) = x^3 - 6x^2 + 9x. f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3). Critical points: x=1, x=3. f''(x) = 6x - 12. f''(1) = -6 < 0 → local max at (1, 4). f''(3) = 6 > 0 → local min at (3, 0). Inflection: f''=0 at x=2. Point: (2, 2). Concave down on (-inf, 2), concave up on (2, inf).
8. Absolute Extrema on a Closed Interval
On [a, b], the absolute maximum and minimum occur at either critical points or endpoints. Procedure: 1. Find f'(x) = 0 on (a, b). 2. Evaluate f at all critical points and at a and b. 3. The largest value is the absolute max; the smallest is the absolute min. Example: Find the absolute extrema of f(x) = x^2 - 4x on [0, 5]. f'(x) = 2x - 4 = 0 at x = 2. f(0) = 0, f(2) = -4, f(5) = 5. Absolute min: -4 at x=2. Absolute max: 5 at x=5.
9. Rolle's Theorem
Rolle's Theorem is a special case of the MVT: If f(a) = f(b) = 0 and f is continuous on [a,b] and differentiable on (a,b), then there exists c in (a,b) with f'(c) = 0. This guarantees that between any two zeros of a smooth function, there is a horizontal tangent — and hence a turning point.
10. Common Mistakes
- Calling f'' = 0 an inflection point without checking that f'' actually changes sign. - Confusing local extrema with absolute extrema. - Forgetting to check endpoints when finding absolute extrema on a closed interval. - Not setting up the sign chart properly and misreading intervals of increase/decrease.