Optimization, Curve Behavior, and the MVT
Core existence theorems and derivative tests that turn local slope information into global statements about functions.
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Questions Covered in This Set
10 cards to master
Extreme Value Theorem (statement)
If f is continuous on a closed, bounded interval [a,b], then f attains an absolute maximum and an absolute minimum on [a,b]. It is an existence theorem — it does not locate the extrema.
Why does EVT fail on (0,1) for f(x)=x?
The interval isn't closed. f gets arbitrarily close to 0 and 1 but never attains them, so no absolute max or min exists. Each EVT hypothesis (closed, bounded, continuous) is necessary.
Fermat's Theorem
If f has a local extremum at an interior point c and f'(c) exists, then f'(c)=0. (The converse is false — f'(c)=0 doesn't guarantee an extremum.)
Sketch the proof of Fermat's Theorem
At a local max: for h>0 the difference quotient [f(c+h)-f(c)]/h ≤ 0, so the right-hand limit ≤ 0; for h<0 it is ≥ 0, so the left-hand limit ≥ 0. Differentiability forces both to equal f'(c), hence f'(c)=0.
What is a critical point?
An interior point where f'(c)=0 or where f'(c) fails to exist (e.g. a cusp or corner).
Closed interval method
To find absolute extrema of continuous f on [a,b]: evaluate f at all points where f'=0, all points where f' doesn't exist, and both endpoints a and b; compare the values.
Rolle's Theorem
If f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then there exists c in (a,b) with f'(c)=0. Proved from EVT + Fermat.
Mean Value Theorem
If f is continuous on [a,b] and differentiable on (a,b), there exists c in (a,b) with f'(c) = [f(b)-f(a)]/(b-a): some instantaneous rate equals the average rate.
How is the MVT derived from Rolle?
Subtract the secant line: let g(x) = f(x) − [f(a) + ((f(b)−f(a))/(b−a))(x−a)]. Then g(a)=g(b)=0, so Rolle gives g'(c)=0, i.e. f'(c) equals the secant slope.
Why does the MVT matter conceptually?
It converts local (pointwise derivative) information into global statements on an interval — e.g. f'>0 implies increasing, f'=0 implies constant — and underlies the Fundamental Theorem of Calculus.
