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That’s the “mean old” value theorem for those of you who can be threatened by it!
The problem with graphing is. . .  Sometimes the graph is misleading! Try graphing  It appears to have a maximum at x = 1.  Evaluate at x = 1.  You will find that f(1) = 0, not 1.  Analytical analysis is necessary to get the whole story
Rolle’s Theorem applies to things “on the level”.  The Mean Value Theorem applies to things that tilt! In other words, the slope of the tangent line for at least one point in the interval HAS to match up to the slope of the secant line between endpoints of the interval!  “  Mean” refers to average rate of change
The Mean Value theorem has some direct use in problem solving, but its main strength is its use in proving other theorems.  The theorem guarantees the existence of a tangent line parallel to the secant line between endpoints. It also implies that there must be a point in the interval (a, b) at which the instantaneous rate of change is equal to the average rate of change over the interval [a, b]
Example 3  p. 175  Finding a tangent line Give f(x) = 5 – (4/x), find all values of c in the open interval (1,4) such that  Solution:  The slope of secant line through (1, f(1)) and (4, f(4)) is  because  the function f  satisfies conditions of mean value theorem, there has to exist at least one number c in interval (1, 4) such that f ’(c)= 1. Set this slope function equal to 1.  So x=   2 In the interval (1,4),  let c = 2
 
Ex 4, p. 175  Finding an instantaneous rate of change Two police cars equipped with radar are parked 5 miles apart on a highway.  As a truck passes the first car, its speed is clocked at 55 miles per hour.  Four minutes later, when it passes the second car, its speed is clocked at 50 miles per hour.  Did the truck exceed the speed limit of 55 miles per hour at some time in that 4 minutes?
Solution: Let t=0 be the time the truck passed the first patrol car.  Then t = 4/60 would be time in hours when it passes the second patrol car. Let s(t) represent position.  Then s(0) = 0 and s(1/15) = 5 Due to mean value theorem (assuming s(t) is differentiable) there has to be a time when truck traveled 75 mph in elapsed time of four minutes.
Alternate form -  if f is continuous on [a, b] and differentiable on (a.b), there exists a number c in (a, b) such that  Assignment p. 176/33-45 EOO, 51-59 odd, 73-75

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Calc 3.2b

  • 1. That’s the “mean old” value theorem for those of you who can be threatened by it!
  • 2. The problem with graphing is. . . Sometimes the graph is misleading! Try graphing It appears to have a maximum at x = 1. Evaluate at x = 1. You will find that f(1) = 0, not 1. Analytical analysis is necessary to get the whole story
  • 3. Rolle’s Theorem applies to things “on the level”. The Mean Value Theorem applies to things that tilt! In other words, the slope of the tangent line for at least one point in the interval HAS to match up to the slope of the secant line between endpoints of the interval! “ Mean” refers to average rate of change
  • 4. The Mean Value theorem has some direct use in problem solving, but its main strength is its use in proving other theorems. The theorem guarantees the existence of a tangent line parallel to the secant line between endpoints. It also implies that there must be a point in the interval (a, b) at which the instantaneous rate of change is equal to the average rate of change over the interval [a, b]
  • 5. Example 3 p. 175 Finding a tangent line Give f(x) = 5 – (4/x), find all values of c in the open interval (1,4) such that Solution: The slope of secant line through (1, f(1)) and (4, f(4)) is because the function f satisfies conditions of mean value theorem, there has to exist at least one number c in interval (1, 4) such that f ’(c)= 1. Set this slope function equal to 1. So x=  2 In the interval (1,4), let c = 2
  • 6.  
  • 7. Ex 4, p. 175 Finding an instantaneous rate of change Two police cars equipped with radar are parked 5 miles apart on a highway. As a truck passes the first car, its speed is clocked at 55 miles per hour. Four minutes later, when it passes the second car, its speed is clocked at 50 miles per hour. Did the truck exceed the speed limit of 55 miles per hour at some time in that 4 minutes?
  • 8. Solution: Let t=0 be the time the truck passed the first patrol car. Then t = 4/60 would be time in hours when it passes the second patrol car. Let s(t) represent position. Then s(0) = 0 and s(1/15) = 5 Due to mean value theorem (assuming s(t) is differentiable) there has to be a time when truck traveled 75 mph in elapsed time of four minutes.
  • 9. Alternate form - if f is continuous on [a, b] and differentiable on (a.b), there exists a number c in (a, b) such that Assignment p. 176/33-45 EOO, 51-59 odd, 73-75