# The Mean Value Theorem Math 131 Spring 2010. Max-Min Theorem If f'(x) is defined on an open interval (a, b) and if f(x) has a relative extrema at a point.

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The Mean Value Theorem Math 131 Spring 2010

Max-Min Theorem If f'(x) is defined on an open interval (a, b) and if f(x) has a relative extrema at a point c on the interval, then f'(c) = 0 Note that f'(c) = 0 is only a necessary and not a sufficient condition; the existence of a zero derivative does not guarantee a relative extrema

Proof (outline) If exists then both the left hand limit and the right hand limit exist Assume that is a local (relative) maximum

In the left hand limit is always less than or equal to since is a relative maximum. Therefore the left hand derivative (?) is always less than or equal to 0.

Using a similar argument one can show that the right hand derivative is always greater than or equal to 0 Therefore since the left and right hand limits exist and equal each other, we conclude that the derivative equals 0 – their “common value”.

Rolle’s Theorem If f’(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) and if f(a) = 0 = f(b) then somewhere on the open interval (a, b) there is a point c such that f‘(c) = 0. This result follows from the Max-Min theorem.

Proof of Rolle’s Theorem Either f(x) is constant on [a, b] which means its derivative f’(x) = 0 everywhere Or starting from f(a) = 0, f(x) increases on the interval (a, b) only to “turn around” (relative maximum) to return to f(b) = 0 Or starting from f(a) = 0, f(x) decreases on the interval (a, b) only to “turn around” (relative minimum) to return to f(b) = 0. In the last two cases the Max-Min theorem says f ‘(c) = 0 at the relative maximum/minimum.

The only real application of Rolle’s Theorem is using it in the proof of the Mean Value Theorem

Mean Value Theorem If f(x) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b) then there is a point c on the open interval (a, b) such that That is, somewhere there is a derivative equal to the average rate of change over the interval

In the next diagram the average rate of change of a function f(x) over the interval [a, b] is seen as the slope of the blue secant line. Observe that we can move the blue secant line “in parallel” so that it touches the curve at a single point c – becoming a (red) tangent line whose slope is f’(c). Thus the Mean Value Theorem

m tan = f’(c) f(x) a c b

Proof: Define: Observe: Therefore by Rolle’s Theorem there is a point c on the interval (a, b) such that Solving for f’(c) yields QED!

Example: is differentiable on (-1, 3). Find the point c where and compute the equation of the tangent line. where. Solving yields Since the equation of the tangent line is

The MVT: Urban Legends A driver enters the State Turnpike at 2:00 PM and at 4:00 PM exits at a point 140 miles away. His average speed is 70 mph. The posted speed limit on the turnpike is 65 mph. Therefore by the Mean Value Theorem, somewhere his speedometer (instantaneous rate of change) equaled his average rate change – 70 mph. The driver was immediately arrested - for speeding!

The MVT: Urban Legends Although there have been rumors that some states have sought to apply the Mean Value Theorem to catch speeders, there is no hard evidence that this was ever done (successfully). See: Snopes.com: E-ZPass Speeding Tickets http://www.snopes.com/autos/law/ezpass.asp

Another Application of the MVT Proving if f’(x) = 0 on an interval then f(x) is a constant function Suppose f’(x) = 0 on some interval I but f(x) is not a constant function. Therefore there are two points a and b on interval I such that f(a) ≠ f(b) Therefore the average rate of change on the interval [a,b] is <- not zero!

Another Application of the MVT Proving if f’(x) = 0 on an interval then f(x) is a constant function (cont.) By the Mean Value Theorem there is a point c on the interval [a,b] such that But wait a minute! f’(c) = 0 so something is wrong! Where did we go wrong? We assumed that f(x) is not a constant function Therefore f(x) is constant on the interval I. QED!

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