Special case of MVT: If f(a) = f(b) = 0 and f is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that f'(c) = 0.

If f is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$

Geometric meaning: Slope of tangent equals slope of secant line at some point.

If f is continuous on closed interval [a, b], then f attains both:

• Absolute maximum value (greatest f(x))
• Absolute minimum value (least f(x))

For critical number c (where f'(c) = 0 or undefined):

• If f' changes from + to - at c -> local maximum
• If f' changes from - to + at c -> local minimum
• If f' does NOT change sign -> no extremum

Critical numbers: Where f'(c) = 0 or f'(c) undefined

Using first derivative test:

1. Test sign of f' on either side of c
2. Determine if f' changes sign

Using second derivative test: If f'(c) = 0:

• f''(c) > 0 -> local minimum
• f''(c) < 0 -> local maximum
• f''(c) = 0 -> test inconclusive

• f''(x) > 0 -> concave up (holds water)
• f''(x) < 0 -> concave down (spills water)
• f''(c) = 0 and sign changes -> inflection point

Relationships:

• Where f is increasing -> f' > 0 (above x-axis)
• Where f is decreasing -> f' < 0 (below x-axis)
• Where f has local max/min -> f' = 0 (x-intercept)
• Where f is concave up -> f'' > 0 (increasing f')
• Where f is concave down -> f'' < 0 (decreasing f')

Strategy:

1. Identify quantity to optimize (maximize or minimize)
2. Write objective function
3. Use constraint to reduce variables
4. Find derivative, set = 0, solve
5. Verify with second derivative test or endpoint analysis

Candidates test for closed interval [a, b]:

1. Find critical numbers in (a, b)
2. Evaluate f at critical numbers
3. Evaluate f at endpoints a and b
4. Largest value = absolute maximum
5. Smallest value = absolute minimum