# Rolle's Thoerem

$f(x)=x^{2}-3x$
$x$
    Factor the expression to obtain $x(x-3)=0$. $x=0$ and $x=3$ are our two endpoints. We know that $f(0)$ and $f(3)$ are the same, thus that satisfies the first part of Rolle's theorem ($f(a)=f(b)$).
2: Now by Rolle's Theorem, we know that somewhere between these points, the slope will be zero. Where? Easy: Take the derivative.
    ${\frac {dy}{dx}}$ $=2x-3$
$x=3/2$
$3/2$
$x$
    Factor the expression to obtain $x(x-3)=0$. $x=0$ and $x=3$ are our two endpoints. We know that $f(0)$ and $f(3)$ are the same, thus that satisfies the first part of Rolle's theorem ($f(a)=f(b)$).
2: Now by Rolle's Theorem, we know that somewhere between these points, the slope will be zero. Where? Easy: Take the derivative.
    ${\frac {dy}{dx}}$ $=2x-3$
$x=3/2$
$3/2$

# Mean Value Theorem

$h(a)=h(b)$
$h(x)$
${\begin{aligned}h(a)&=f(a)(g(b)-g(a))-g(a)(f(b)-f(a)-f(a)g(b)+f(b)g(a)\\&=f(a)g(b)-f(a)g(a)-g(a)f(b)-g(a)f(a)-f(a)g(b)+f(b)g(a)\\&=0\end{aligned}}$
${\begin{aligned}h(b)&=f(b)(g(b)-g(a))-g(b)(f(b)-f(a))-f(a)g(b)+f(b)g(a)\\&=f(b)g(b)-f(b)g(a)-g(b)f(b)-g(b)f(a)-f(a)g(b)+f(b)g(a)\\&=0\end{aligned}}$
${\begin{aligned}h(a)&=f(a)(g(b)-g(a))-g(a)(f(b)-f(a)-f(a)g(b)+f(b)g(a)\\&=f(a)g(b)-f(a)g(a)-g(a)f(b)-g(a)f(a)-f(a)g(b)+f(b)g(a)\\&=0\end{aligned}}$
${\begin{aligned}h(b)&=f(b)(g(b)-g(a))-g(b)(f(b)-f(a))-f(a)g(b)+f(b)g(a)\\&=f(b)g(b)-f(b)g(a)-g(b)f(b)-g(b)f(a)-f(a)g(b)+f(b)g(a)\\&=0\end{aligned}}$
$g(x)=x$
$g'(x)=1$
$g(b)-g(a)=b-a$
$b\neq a$
${\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}$
$f'(c)={\frac {f(b)-f(a)}{b-a}}$
$g(x)=x$
$g'(x)=1$
$g(b)-g(a)=b-a$
$b\neq a$
${\frac {f'(c)}{g'(c)}}={\frac {f(b)-f(a)}{g(b)-g(a)}}$
$f'(c)={\frac {f(b)-f(a)}{b-a}}$
$x=c$
$f(x)=x^{3}$
$x=0$
$x=2$
    ${\frac {f(b)-f(a)}{b-a}}$
insert values. Our chosen interval is $[0,2]$. So, we have
    ${\frac {f(2)-f(0)}{2-0}}={\frac {8}{2}}=4$
2: By the Mean Value Theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point $x=c$.
    ${\frac {dy}{dx}}=3x^{2}$
$c$
$4=3x^{2}$
$x={\sqrt {4/3}}=\mathbf {\frac {2{\sqrt {3}}}{3}}$
    ${\frac {f(b)-f(a)}{b-a}}$
insert values. Our chosen interval is $[0,2]$. So, we have
    ${\frac {f(2)-f(0)}{2-0}}={\frac {8}{2}}=4$
2: By the Mean Value Theorem, we know that somewhere in the interval exists a point that has the same slope as that point. Thus, let us take the derivative to find this point $x=c$.
    ${\frac {dy}{dx}}=3x^{2}$
$c$
$4=3x^{2}$
$x={\sqrt {4/3}}=\mathbf {\frac {2{\sqrt {3}}}{3}}$
$f(x)=\sin(x)$
$[0,\pi ]$
    ${\frac {f(b)-f(a)}{b-a}}$
so,
    ${\frac {\sin(\pi )-\sin(0)}{\pi -0}}=0$
(Remember, sin(π) and sin(0) are both 0.)
2: Now that we have the slope of the line, we must find the point x = c that has the same slope. We must now get the derivative!
    ${\frac {d\sin(x)}{dx}}=\cos(x)=0$
$\pi /2+\pi n$
$n$
$[0,\pi ]$
$\mathbf {\pi /2}$
$c$
    ${\frac {f(b)-f(a)}{b-a}}$
so,
    ${\frac {\sin(\pi )-\sin(0)}{\pi -0}}=0$
(Remember, sin(π) and sin(0) are both 0.)
2: Now that we have the slope of the line, we must find the point x = c that has the same slope. We must now get the derivative!
    ${\frac {d\sin(x)}{dx}}=\cos(x)=0$
$\pi /2+\pi n$
$n$
$[0,\pi ]$
$\mathbf {\pi /2}$
$c$
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