Calc2.65

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We wish to show that there is a c in (a,b) such that f'(c)={\frac  {f(b)-f(a)}{b-a}}

From Rolle's theorem we have that if f(a)=f(b), then there is a c in (a,b) such that f'(c)=0

Consider h(x)=[b-a]f(x)-[f(b)-f(a)]x\,

h(b)=[b-a]f(b)-[f(b)-f(a)]b=bf(a)-af(b)\,

h(a)=[b-a]f(a)-[f(b)-f(a)]a=bf(a)-af(b)\,

So h(a)=h(b)

We can apply Rolle's Theorem to h:
h'(c)=[b-a]f'(c)-[f(b)-f(a)]=0\,

\Rightarrow f'(c)={\frac  {f(b)-f(a)}{b-a}}