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. b a
DefinitionLocal 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 Max Local Min Local Min
Absolute and Local Max (a, f(a)) (c, f (c)) Absolute max on [a,b] (b, f(b)) Local Max a c b
Critical PointsDefinition 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 IntervalTheorem: 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.
ExampleFind 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 Endpoint Max (6, 43) Critical Point – local max (1, 18) Endpoint (0, 7) Critical point – local min (4, -9)
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(c) f(a) f(b) a c b
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).