Local Extrema & Mean Value Theorem Rolle’s theorem: What goes up must come down Mean value theorem: Average velocity must be attained Some applications: Inequalities, Roots of Polynomials

Local Extrema

Example Let f(x) = (x-1)2(x+2), -2  x  3 Use the graph of f(x) to find all local extrema Find the global extrema

Example Consider f(x) = |x2-4| for –2.5  x < 3 Find all local and global extrema

Fermat’s Theorem Theorem: If f has a local extremum at an interior point c and f (c) exists, then f (c) = 0. Proof Case 1: Local maximum at interior point c Then derivative must go from 0 to 0 around c Proof Case 2: Local minimum at interior point c Then derivative must go from 0 to 0 around c

Cautionary notes f (c) = 0 need not imply local extrema Function need not be differentiable at a local extremum (e.g., earlier example |x2-4|) Local extrema may occur at endpoints

Summary: Guidelines for finding local extrema: Don’t assume f (c) = 0 gives you a local extrema (such points are just candidates) Check points where derivative not defined Check endpoints of the domain These are the three candidates for local extrema Critical points: points where f (c)=0 or where derivative not defined

What goes up must come down Rolle’s Theorem: Suppose that f is differentiable on (a,b) and continuous on [a,b]. If f(a) = f(b) = 0 then there must be a point c in (a,b) where f (c) = 0.

Proof of Rolle’s theorem If f = 0 everywhere it’s easy Assume that f > 0 somewhere (case f< 0 somewhere similar) Know that f must attain a maximum value at some point which must be a critical point as it can’t be an endpoint (because of assumption that f > 0 somewhere). The derivative vanishes at this critical point where maximum is attained.

Need all hypotheses Suppose f(x) = exp(-|x|). f(-2) = f(2) Show that there is no number c in (-2,2) so that f (c)=0 Why doesn’t this contradict Rolle’s theorem

Examples f(x) = sin x on [0, 2] exp(-x2) on [-1,1] Any even continuous function on [-a,a] that is differentiable on (-a,a)

Average Velocity Must be Attained Mean Value Theorem: Let f be differentiable on (a,b) and continuous on [a,b]. Then there must be a point c in (a,b) where

Idea in Mean Value Proof Says must be a point where slope of tangent line equals slope of secant lines joining endpoints of graph. A point on graph furthest from secant line works:

Consequences f is increasing on [a,b] if f (x) > 0 for all x in (a,b) f is decreasing if f (x) < 0 for all x in (a,b) f is constant if f (x)=0 for all x in (a,b) If f (x) = g (x) in (a,b) then f and g differ by a constant k in [a,b]

Example

More Consequences of MVT

Examples Show that f(x) = x3+2 satisfies the hypotheses of the Mean Value Theorem in [0,2] and find all values c in this interval whose existence is guaranteed by the theorem. Suppose that f(x) = x2 – x –2, x in [-1,2]. Use the mean value theorem to show that there exists a point c in [-1,2] with a horizontal tangent. Find c.

Existence/non-existence of roots x3 + 4x - 1 = 0 must have fewer than two solutions The equation 6x5 - 4x + 1 = 0 has at least one solution in the interval (0,1)