Mean Value Theorem

mean value theorem relate the value of a function to a value of it derivatives. to be accurately, this theorem state that the tangent and the secant line are parallel for a function.

let f(x) be a function. it is continuous  on the close interval [a, b] and differential on the open interval (a, b), then there exist at list one number in the center of the open interval (a, b)

$$\begin{gathered} f\left(\alpha \right) = \frac{f\left(b\right) – f\left(a\right)}{b –a } \\ f\left(b\right) – f\left(a\right) = f‘\left(\alpha \right) (b–a) \\ g\left(x\right) = f\left(a\right) +\left[\frac{f\left(b\right)–f\left(a\right)}{b–a}\right] (x–a) \\ \sin ce the line is a straight line, by inspection g\left(a\right) = f\left(a\right) and g\left(a\right) = f\left(a\right) \\ (f–g)\left(a\right) = f\left(a\right) – g\left(a\right) = f\left(b\right) –g\left(b\right) \\ = (f–g)\left(b\right) \\ thus there exist a point \alpha in the open interval (a, b) such that \\ f(–g) a =0 \\ f–g\left(x\right) – \left[\frac{{\displaystyle f\left(b\right)–f\left(a\right)}}{{\displaystyle b–a}}\right] \\ at x=\alpha the \tan gent line become \\ (f–g) \left(\alpha \right) =0 \\ f\left(\alpha \right) – \left[\frac{{\displaystyle f\left(b\right)–f\left(a\right)}}{{\displaystyle b–a}}\right] =0 \\ f\left(\alpha \right) = \left[\frac{{\displaystyle f\left(b\right)–f\left(a\right)}}{{\displaystyle b–a}}\right] \\ therefore, it can be concluded that the \tan gent line at \alpha and slope of [a, b] are the same. \end{gathered}$$

Example 

$$\begin{gathered} a function f\left(x\right) = x^{3 } on the close interval [2, 4] then, the means value theorem \\ satisfies the folowing; \\ 1. f\left(x\right) = x^3 in continous on the close interval [2, 4] \\ 2. f\left(x\right) = x^3 is differentiable on the open interval (2, 4) \\ then, there is a number \alpha , such that \epsilon (2, 4) and \\ f‘\left(\alpha \right) = \left[\frac{f(5{)}^3 – f(4{)}^3}{5–4}\right] \\ \frac{61}{1} \\ answer is 61 \end{gathered}$$