Difference between revisions of "Lagrange's Mean Value Theorem"

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[https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT]

Revision as of 16:01, 19 February 2024

Lagrange's mean value theorem (often called "the mean value theorem," and abbreviated MVT or LMVT) is considered one of the most important results in real analysis. An elegant proof of the Fundamental Theorem of Calculus can be given using LMVT.

Statement

Let $f:[a,b]\rightarrow\mathbb{R}$ be a continuous function, differentiable on the open interval $(a,b)$. Then there exists some $c\in (a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$.

Informally, this says that a differentiable function must at some point grow with instantaneous velocity equal to its average velocity over an interval.

Proof

We reduce the problem to Rolle's theorem by using an auxiliary function.

Consider $g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a).$ Note that $g(a)=g(b)=f(a).$ By Rolle's theorem, there exists $c$ in $(a,b)$ such that $g'(c)=0,$ or $\[f'(c)-\frac{f(b)-f(a)}{b-a}=0,\] which simplifies to \[f'(c)=\frac{f(b)-f(a)}{b-a},\] as desired.

QED

See Also

TOTO SLOT