Problem of the Day

Problem of the Day If f(x) = x√2x - 3, then f '(x) = A) 3x - 3 √2x - 3 D) -x + 3 √2x - 3 B) x √2x - 3 E) 5x - 6 2√2x - 3 C) 1 √2x - 3

Extreme Value Theorem says ?

Extreme Value Theorem says ? A continuous function on a closed interval [a, b] must have a minimum and a maximum (but they can be at endpoints). It gives the conditions that guarantee an extreme value in the interior of a closed interval.

Rolle's Theorem Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b) then there is at least one number c in (a, b) such that f '(c) = 0. relative max a c b a c b not differentiable so cannot use Rolle's Theorem but it still has a critical number Rolle's Theorem

Example f(x) = x4 - 2x2 + 4 Find all values of c in the interval [-2, 2] where f '(c) = 0.

Example f(x) = x4 - 2x2 + 4 Find all values of c in the interval [-2, 2] where f '(c) = 0. 1) it is continuous 2) it is differentiable on (-2, 2) 3) f(-2) = f(2) 4) use Rolle's Theorem

Example f(x) = x4 - 2x2 + 4 Find all values of c in the interval [-2, 2] where f '(c) = 0. f '(x) = 4x3 - 4x 0 = 4x(x2 - 1) 0 = 4x(x - 1)(x + 1) x = 0, 1, -1

A puzzle to solve A cow stands up, trots around the barn and lays back down. What does Rolle's Theorem have to say about the cow's acceleration during the trip?

A puzzle to solve A cow stands up, trots around the barn and lays back down. What does Rolle's Theorem have to say about the cow's acceleration during the trip? The cow's velocity starts at zero and ends at zero so somewhere during the trip the derivative of velocity (acceleration) of "cow" must be 0.

Mean Value Theorem If f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that f '(c) = f(b) - f(a) b - a What does this really say?

Mean Value Theorem If f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c in (a, b) such that f '(c) = f(b) - f(a) b - a The theorem guarantees a tangent line that is parallel to the secant line through (a, f(a)) and (b, f(b)) The theorem guarantees that there must be a point in the open interval (a, b) where the instantaneous rate of change equals the average rate of change over [a, b].

Example f(x) = x(x2 - x - 2) Find c such that f '(c) = f(b) - f(a) on [1, -1] b - a

Example f(x) = x(x2 - x - 2) Find c such that f '(c) = f(b) - f(a) on [1, -1] b - a 1) find slope of secant line f(1) - f(-1) = -2 - 0 = -2 = -1 1 - (-1) 2 2

Example f(x) = x(x2 - x - 2) Find c such that f '(c) = f(b) - f(a) on [1, -1] b - a 1) find slope of secant line f(1) - f(-1) = -2 - 0 = -2 = -1 1 - (-1) 2 2 2) find x where slope of tangent line is -1

f(x) = x(x2 - x - 2) Example Find c such that f '(c) = f(b) - f(a) on [1, -1] b - a f(1) - f(-1) = -2 - 0 = -2 = -1 1 - (-1) 2 2 1) find slope of secant line 2) find x where slope of tangent line is -1 a) find derivative (i.e. equation for slope of tangent line) f(x) = x(x2 - x - 2) f(x) = x3 - x2 - 2x f '(x) = 3x2 - 2x - 2 b) set derivative equal to -1

f(x) = x(x2 - x - 2) Example Find c such that f '(c) = f(b) - f(a) on [1, -1] b - a f(1) - f(-1) = -2 - 0 = -2 = -1 1 - (-1) 2 2 1) find slope of secant line 2) find x where slope of tangent line is -1 a) find derivative (i.e. equation for slope of tangent line) f '(x) = 3x2 - 2x - 2 b) set derivative equal to -1 -1 = 3x2 - 2x - 2 0 = 3x2 - 2x - 1 0 = (3x + 1)(x - 1) x = -1/3 or 1

Look at example 4 on page 175

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Problem of the Day