AP Calculus4.2 Extreme Values

Applications of the Derivative

•One of the most common applications of the derivative is to find maximum and/or minimum values of a function

•These are called “Extreme Values” or “Extrema”

•Extrema would be an excellent name for an 80’s Hair Band.

Definition

•Let f (x) be a function defined on an interval I, let a ϵ I, then f (a) is

•Absolute minimum of f (x) on I, if f (a) ≤ f (x) for all x in I

•Absolute maximum of f (x) on I, if f (a) > f (x) for all x in I

•If no interval is indicated, then the extreme values apply to the entire function over its domain.

Do All Functions Have Extrema?• f (x) = x•No extrema unless the function is defined

on an interval.

Do All Functions Have Extrema?•g (x) = (-x(x2 – 4))/x•Discontinuous and has no max on [a, b]

a b

Do All Functions Have Extrema?• f (x) = tan x•No max or min on the open interval (a, b)

a b

Do All Functions Have Extrema?•h(x) = 3x3 + 6x2 + x + 3•Function is continuous and [a, b] is

closed. Function h(x) has a min and max.

a

b

Definition

Local Extrema a function f (x) has a:•Local Minimum at x = c if f (c) is the

minimum value of f on some open interval (in the domain of f) containing c.

•Local Maximum at x = c if f (c) is the maximum value of f on some open interval containing c.

Local Max and Min

Local Min

Local Max

Local Min

Absolute and Local Max

a b

Absolute max on [a,b]

(a, f(a))

(b, f(b))Local Max

c

(c, f (c))

Critical Points

Definition of Critical Points•A number c in the domain of f is called a

critical point if either f’ (c) = 0 or f’ (c) is undefined.

Fermat’s Theorem

•Theorem: If f (c) is a local min or max, then c is a critical point of f.

•Not all critical points yield local extrema. “False positives” can occur meaning that f’(c) = 0 but f(c) is not a local extremum.

Fermat’s Theorem

f(x) = x3 + 4

Tangent line at (0, 4) is horizontal

f(0) is NOT an extremum

Optimizing on a Closed Interval

Theorem: Extreme Values on a Closed Interval

•Assume f (x) is continuous on [a, b] and let f(c) be the minimum or maximum value on [a, b]. Then c is either a critical point or one of the endpoints a or b.

Example

Find the extrema of f(x) = 2x3 – 15x2 + 24x + 7 on [0, 6].

•Step 1: Set f’(x) = 0 to find critical points▫f’(x) = 6x2 – 30x + 24 = 0, x = 1, 4

•Step 2: Calculate f(x) at critical points and endpoints.▫f(1) = 18, f(4) = -9, f(0) = 7, f(6) = 43

•The maximum of f(x) on [0, 6] is (6, 43) and minimum is (4, -9).

Graph of f(x)=2x3-15x2+24x+7

EndpointMax (6, 43)

Critical Point – local max (1, 18)

Critical point – local min (4, -9)

Endpoint(0, 7)

Example

•Compute critical points of•Find extreme values on [0, 2]

Example

•Critical point: h’(t) = 0, t = 0•Local min (0, -1)•Endpoints: (-2, 1.44), (2, 1.44) maximums

Example

•Find the extreme values of g(x) = sin x cos x on [0, π]

Example

•Critical points: g’(x) = cos 2 x – sin 2 x•g’(x) = 0, x = π/4, 3π/4•g(π/4) = ½ , max•g(3π/4) = -1/2 , min•Endpoints (0, 0), (π, 0)

Rolle’s Theorem

•Assume f (x) is continuous on [a, b] and differentiable on (a, b). If f (a) = f (b) then there exists a number c between a and b such that f’(c) = 0

f(a) f(b)

a bc

f(c)

Example

•Use Rolle’s Theorem to show that the function f(x) = x3 + 9x – 4 has at most 1 real root.

Example

•If f (x) had 2 real roots a and b, then f (a) = f (b) and Rolle’s Theorem would apply with a number c between a and b such that f’(c) = 0.

•However…f’(x) = 3x2 + 9 and 3x2 + 9 = 0 has no real solutions, so there cannot be a value c such that f’ (c) = 0 so there is not more than 1 real root of f (x).