Main Concept

The Mean Value Theorem (MVT) states that if a function $f$ is continuous on the closed interval $\left[a\&comma;b\right]$ and differentiable on the open interval $\left(a\&comma;b\right)$ where $a<b$, then there exists a point $c$ in $\left(a\&comma;b\right)$ such that $f \&apos;\left(c\right)\&equals;\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

In other words, for a function which changes smoothly over an interval, there must be at least one point in the interval where the instantaneous rate of change of the function equals its average rate of change over the whole interval. In terms of the graph of the function, if the graph is smooth, then there must be some point between the two endpoints where the tangent at that point is parallel to the secant line connecting the two endpoints.

As can be seen in the proof below, the Mean Value Theorem follows from a specific statement of Rolle's Theorem: if a real-valued function $f$  is continuous on the closed interval $\left[a\&comma;b\right]$, differentiable on the open interval $\left(a\&comma;b\right)$, and $f\left(a\right) \&equals; f\left(b\right)$, then there exists a $c$ in $\left(a\&comma;b\right)$ such that $f \&apos;\left(c\right)\&equals;0$.

The MVT is one of the most important theorems in differential calculus, as it is essential in proving the Fundamental Theorem of Calculus and the general statement of Taylor's Theorem (with the Lagrange form of the remainder term).

Proof of the Mean Value Theorem

Consider the function $h\left(x\right) \= f\left(x\right)\cdot \left(b-a\right) - x\cdot \left(f\left(b\right) - f\left(a\right)\right)$, which is continuous on $\left[a\,b\right]$ because $f\left(x\right)$ is continuous on this interval. Taking the derivative, we obtain: $h \'\left(x\right) \= f \'\left(x\right)\cdot \left(b-a\right) - \left(f\left(b\right) - f\left(a\right)\right)$ which is differentiable on $\left(a\, b\right)$ because $f\left(x\right)$ is differentiable on this interval. Note that: $h\left(a\right) \= f\left(a\right)\cdot \left(b - a\right) - a\cdot \left(f\left(b\right) - f\left(a\right)\right)$ $\= b\cdot f\left(a\right) - a\cdot f\left(a\right) - a\cdot f\left(b\right) \+ a\cdot f\left(a\right)$                         $\= b\cdot f\left(a\right) - a\cdot f\left(b\right)$                         $\= b\cdot f\left(b\right) - a\cdot f\left(b\right) - b\cdot f\left(b\right) \+ b\cdot f\left(a\right)$                         $\= f\left(b\right)\cdot \left(b-a\right) -b\cdot \left(f\left(b\right) - f\left(a\right)\right)$                         $\= h\left(b\right)$ Therefore by Rolle's Theorem, there exists a point $c$ in $\left(a\, b\right)$ such that: $h \'\left(c\right) \= 0 \= f \'\left(c\right)\cdot \left(b-a\right) - \left(f\left(b\right) - f\left(a\right)\right)$. And so, $f\left(b\right) - f\left(a\right) \= f \'\left(c\right)\cdot \left(b-a\right)$ which can be rearranged to give us $f \'\left(c\right) \= \frac{f\left(b\right) - f\left(a\right)}{b - a}$.

Interval

$\le x \le$

Current value of $x$ :

Slope of the secant:

Slope of the tangent:

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