MHB Mean Value Theorem: Showing Change in a Function is Bounded

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SUMMARY

The discussion centers on the Mean Value Theorem (MVT) and its application to the function $f(x)$ over the interval $[2,8]$. It establishes that the average rate of change of the function is calculated as $\dfrac{f(8)-f(2)}{8-2}$. According to the MVT, there exists at least one value of $x$ in the interval $(2,8)$ such that $f'(x)$ equals this average rate of change. The bounds on the derivative, $3 \le f'(x) \le 5$, lead to the conclusion that $18 \le f(8)-f(2) \le 30$.

PREREQUISITES
  • Understanding of the Mean Value Theorem (MVT)
  • Basic calculus concepts, including derivatives
  • Knowledge of function notation and intervals
  • Ability to perform algebraic manipulation of inequalities
NEXT STEPS
  • Study the implications of the Mean Value Theorem in different contexts
  • Explore examples of functions that satisfy the conditions of the MVT
  • Learn about the relationship between derivatives and the behavior of functions
  • Investigate applications of the MVT in real-world scenarios
USEFUL FOR

Students studying calculus, educators teaching mathematical concepts, and anyone interested in the application of the Mean Value Theorem in analyzing function behavior.

karush
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Ok Just have trouble getting this without a function..
 
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average rate of change of $f(x)$ on the interval $[2,8]$ is $\dfrac{f(8)-f(2)}{8-2}$

the MVT states there exists at least one value of $x \in (2,8)$ where $f'(x) = \dfrac{f(8)-f(2)}{8-2}$

$3 \le f'(x) \le 5 \implies 3 \le \dfrac{f(8)-f(2)}{8-2}\le 5 \implies 18 \le f(8)-f(2) \le 30$
 

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