Mean Value Theorem: Solving for f(8) -f(2)

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SUMMARY

The discussion focuses on applying the Mean Value Theorem (MVT) to determine the bounds for the difference f(8) - f(2) given the constraints on the derivative f'(x). Specifically, it establishes that if 3 ≤ f'(x) ≤ 5 for all x, then 18 ≤ f(8) - f(2) ≤ 30. The MVT states that there exists a point c in the interval [2, 8] such that f'(c) equals the average rate of change over that interval.

PREREQUISITES
  • Understanding of the Mean Value Theorem (MVT)
  • Basic knowledge of calculus, specifically derivatives
  • Ability to manipulate inequalities
  • Familiarity with function notation and evaluation
NEXT STEPS
  • Review the Mean Value Theorem and its applications in calculus
  • Practice solving problems involving bounds on function differences using derivatives
  • Explore examples of applying MVT to different functions
  • Study the implications of derivative constraints on function behavior
USEFUL FOR

Students studying calculus, particularly those learning about the Mean Value Theorem and its applications in determining function behavior based on derivative constraints.

susan__t
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Homework Statement


Suppose that 3 is < and equal than f'(x) which is also < and equal to 5 for all vales of x. Show that 18< or equal to f(8) -f(2) < or equal to 30.


Homework Equations


Mean Value theorem


The Attempt at a Solution


I have no clue where to start or which values associate with the values of those of MVT
 
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This one looks pretty straightforward: What does the mean value theorem say about [tex]\frac{f(8)-f(2)}{8-2}[/tex]?
 
Start with the MVT, couched in terms of your problem. It says that for some c, with c in [?, ?], f'(c) = ?. Then, what other information are you given?
 

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