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Let $f$ be three times differentiable on $[a,b]$. $f'''$ is continuous. Show that there is a $c\in[a,b]$ such that $$f(b)-f(a) =((b-a)/2)\cdot(f'(a)+f'(b))-((b-a)³/12)\cdot f'''(c)$$

This looks like a combination of Mean Value Theorem and Taylor's series. but how to solve it, i don't know. Thanks for any answers!

Paramanand Singh
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alf
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  • There is a typo. You must have $f'(a) + f'(b)$ instead of $f'(a) - f'(b)$. Check with $f(x) = x$. And then it is same question as http://math.stackexchange.com/questions/1008049/approximating-a-three-times-differentiable-function-by-a-linear-combination-of-d – Paramanand Singh Nov 14 '14 at 06:17
  • I have edited the question as per my previous comment. – Paramanand Singh Nov 14 '14 at 06:20

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Hint: Use $$ f(x)=f(a)+f'(a)(x-a)+\frac{1}{2}f''(a)(x-a)^2+\frac{1}{6}f'''(c_1)(x-a)^3$$ and $$ f(x)=f(b)+f'(b)(x-a)+\frac{1}{2}f''(b)(x-b)^2+\frac{1}{6}f'''(c_2)(x-b)^3$$ for $x=b$ and $x=a$ respectively.

xpaul
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