Most important in differential calculus: Optimization Problems

1. global (not local! end point) max Extreme Value Theorem: If f is continuous on [a,b] then f attains abs maxf(c) and abs min f(d) at some c, d[a,b]. not global, not local local max local min global AND local min Most important in differential calculus:Optimization Problems Maximum & Minimum values f has absolute (global) maximum at the point c, if f(c) f(x) xD – domain of f. extreme values f has absolute (global) minimum at the point c, if f(c) f(x) xD – domain of f. f has local (relative) maximum at the point c, if f(c) f(x) x near c, (i.e. x(c-e,c+e), e>0). f has local (relative) minimum at the point c, if f(c)  f(x) x near c, (i.e. x(c-e,c+e), e>0). y y f(1)  3 but f(1)3 no maximum: not continuous no maximum, no minimum: open interval 3 1 1 0 1 2 x 0 1 2 x

2. y f(c)=0 f(d)=0 0 c d x Fermat’s Theorem: If f has a local maximum or minimum at c then c is a critical number of f. Fermat’s Theorem: If f has a local maximum or minimum at c and if f(c) exists, then f(c)=0. How to find local extreme values? Butf(c)=0 doesn’t imply that c is local max or min! (e.g. f(x)=x3, c=0). Existense of f(c) is important (e.g. f(x)=|x|, c=0). To find local extreme values: start from c, s.t. f(c)=0 or f(c) doesn’t exist. c – critical number of f. local max/min, i.e. at critical number or Absolute max/min of continuous function f on [a,b] = at endpoint of the interval • How to find absolute max/min? • find critical numbers in (a,b) and values of f there; • find values of f at endpoints; • abs max=max(step1,step2), abs min=min(step1,step2).

3. Proof of Fermat’s th. Assume f has a local max at c for all x near c, i.e. for any small h (both positive and negative), Divide by h: If h>0 Since f(c) exists, on the other hand, If h<0 Since f(c) exists, Therefore, If f has a local min at c, the proof is similar.

4. Rolle’s Theorem: Let f be • continuous on [a,b]; • differentiable on (a,b); • f(a)=f(b). •   c(a,b), s.t. f(c)=0. f(c)=0 f(c)=0 f(c)=0 f(d)=0 f(c)=0 a c b a c b a c d b a c b f(a) f(b) f(b) f(a) a c b a c d b Th: If f(x)=0 x(a,b)then f is constant on (a,b). Corollary: If f(x)=g(x) x(a,b)then f -g is constant on (a,b). Case 1 Case 2 Case 3 Ex. Show that f(x)=1+2x+x3+4x5 has exactly one real root. • Mean Value Theorem: Let f be • continuous on [a,b]; • differentiable on (a,b). •  c(a,b), s.t. or

5. Proof of Rolle’s th. Case 1.f(x) is constant Case 2. there exists By Extreme Value th., f has an absolute max in [a,b]. Since f(a)=f(b) is not a maximum value, abs max is achieved at c(a,b), i.e. c – local max. f is differentiable at c  by Fermat’s th., Case 3. similar to Case 2: there exists By Extreme Value th., f has an absolute min in [a,b]. Since f(a)=f(b) is not a minimum value, abs min is achieved at c(a,b), i.e. c – local min. f is differentiable at c  by Fermat’s th.,

6. If f(x)>0 on interval I, then f is increasing on I. If f(x)<0 on interval I, then f is decreasing on I. • First derivative test. Let f be a continuous function, c – its critical number. • If f changes from positive to negative at c, then f has local max at c; • If f changes from negative to positive at c, then f has local min at c; • If f doesn’t change sign at c, then f has no local max or min at c. If f(x)>0 on interval I, then graph is concave upward on I. If f(x)<0 on interval I, then graph is concave downward on I. • Second derivative test. If f is continuous near c. • If f(c)=0 and f(c)>0, then f has local min at c; • If f(c)=0 and f(c)<0, then f has local max at c; Concave upward Concave downward