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)

$\phantom{\rule{0ex}{0ex}}f\left(\alpha \right)=\frac{f\left(b\right)\u2013f\left(a\right)}{b\u2013a}\phantom{\rule{0ex}{0ex}}f\left(b\right)\u2013f\left(a\right)=f\u2018\left(\alpha \right)(b\u2013a)\phantom{\rule{0ex}{0ex}}g\left(x\right)=f\left(a\right)+\left[\frac{f\left(b\right)\u2013f\left(a\right)}{b\u2013a}\right](x\u2013a)\phantom{\rule{0ex}{0ex}}\mathrm{sin}cethelineisastraightline,byinspectiong\left(a\right)=f\left(a\right)andg\left(a\right)=f\left(a\right)\phantom{\rule{0ex}{0ex}}(f\u2013g)\left(a\right)=f\left(a\right)\u2013g\left(a\right)=f\left(b\right)\u2013g\left(b\right)\phantom{\rule{0ex}{0ex}}=(f\u2013g)\left(b\right)\phantom{\rule{0ex}{0ex}}thusthereexistapoint\alpha intheopeninterval(a,b)suchthat\phantom{\rule{0ex}{0ex}}f(\u2013g)a=0\phantom{\rule{0ex}{0ex}}f\u2013g\left(x\right)\u2013\left[\frac{{\displaystyle f\left(b\right)\u2013f\left(a\right)}}{{\displaystyle b\u2013a}}\right]\phantom{\rule{0ex}{0ex}}atx=\alpha the\mathrm{tan}gentlinebecome\phantom{\rule{0ex}{0ex}}(f\u2013g)\left(\alpha \right)=0\phantom{\rule{0ex}{0ex}}f\left(\alpha \right)\u2013\left[\frac{{\displaystyle f\left(b\right)\u2013f\left(a\right)}}{{\displaystyle b\u2013a}}\right]=0\phantom{\rule{0ex}{0ex}}f\left(\alpha \right)=\left[\frac{{\displaystyle f\left(b\right)\u2013f\left(a\right)}}{{\displaystyle b\u2013a}}\right]\phantom{\rule{0ex}{0ex}}therefore,itcanbeconcludedthatthe\mathrm{tan}gentlineat\alpha andslopeof[a,b]arethesame.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Example

$afunctionf\left(x\right)={x}^{3}onthecloseinterval[2,4]then,themeansvaluetheorem\phantom{\rule{0ex}{0ex}}satisfiesthefolowing;\phantom{\rule{0ex}{0ex}}1.f\left(x\right)={x}^{3}incontinousonthecloseinterval[2,4]\phantom{\rule{0ex}{0ex}}2.f\left(x\right)={x}^{3}isdifferentiableontheopeninterval(2,4)\phantom{\rule{0ex}{0ex}}then,thereisanumber\alpha ,suchthat\epsilon (2,4)and\phantom{\rule{0ex}{0ex}}f\u2018\left(\alpha \right)=\left[\frac{f(5{)}^{3}\u2013f(4{)}^{3}}{5\u20134}\right]\phantom{\rule{0ex}{0ex}}\frac{61}{1}\phantom{\rule{0ex}{0ex}}answeris61\phantom{\rule{0ex}{0ex}}$

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