Extrema, the Mean Value Theorem, and Extreme Value Theorem
Critical points (where f'=0 or is undefined) are candidates for local extrema. The Extreme Value Theorem guarantees absolute extrema on closed intervals; the Mean Value Theorem guarantees a tangent parallel to a secant line.
Finding extrema
A critical point occurs where f'(x) = 0 or f'(x) is undefined (but f(x) itself is defined). Local extrema can only occur at critical points, but not every critical point is an extremum.
Extreme Value Theorem: if f is continuous on a closed interval [a,b], f is guaranteed to have both an absolute maximum and absolute minimum on that interval. To find them, compare f at all critical points inside [a,b] AND at the two endpoints — the largest and smallest values win.
The Mean Value Theorem (MVT)
If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c in (a,b) where f'(c) = [f(b)-f(a)] / (b-a). In words: somewhere in the interval, the instantaneous rate of change matches the average rate of change over the whole interval.
Both continuity on the closed interval AND differentiability on the open interval are required hypotheses — MVT can fail to guarantee a result if either is violated (e.g., a corner point inside the interval).
Worked Example
Verify MVT applies to f(x) = x² on [1,3] and find the guaranteed value of c.
- f(x)=x² is a polynomial, so it's continuous on [1,3] and differentiable on (1,3) — hypotheses satisfied.
- Average rate of change: [f(3)-f(1)]/(3-1) = (9-1)/2 = 4.
- f'(x) = 2x. Set 2c = 4, so c = 2, which is indeed inside (1,3).
Answer: c = 2
Key terms (2)
- Critical point
- A point where f'(x)=0 or f'(x) is undefined (while f(x) is defined).
- Mean Value Theorem
- Guarantees a point where the instantaneous rate of change equals the average rate of change over [a,b].
Practice Quiz
f(x) = x³ has f'(x) = 3x², so f'(0) = 0. Is x=0 a local extremum?