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Define $h(x)$ on $[a,b]$ by $$ h(x) = f(x) - f(a) - \left( \frac{f(b)-f(a)}{b-a} \right) (x-a) $$ Clearly, $h$ is continuous on $[a,b]$ , differentiable on $(a, b)$ , and
Notice that $h$ satisfies the conditions of Rolle's Theorem. Therefore, by Rolle's Theorem there exists $c \in (a,b)$ such that $h'(c)=0$ .
However, from the definition of $h$ we obtain by differentiation that $$ h'(x) = f'(x) - \frac{f(b)-f(a)}{b-a} $$ Since $h'(c)=0$ , we therefore have $$ f'(c) = \frac{f(b)-f(a)}{b-a} $$ as required.
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- Michael Spivak, Calculus, 3rd ed., Publish or Perish Inc., 1994.
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