# Answer to Question #2398 in Real Analysis for Tom

Question #2398

Suppose f is continuously differentiable on [0,1] and f'' (greater than or equal to) 0 on [0,1]. Prove that f(x) (greater than or equal to) f(c) + f'(c)(x-c) for every x,c in [0,1].

Expert's answer

By Taylor theorem in integral from

<img src="/cgi-bin/mimetex.cgi?f%28c%29=f%28x%29+f%27%28c%29%28x-c%29+%5Cint_c%5Ex%20%5Cfrac%7Bf%27%27%28t%29%7D%7B2%7D%28t-c%29%5E2dt" title="f(c)=f(x)+f'(c)(x-c)+\int_c^x \frac{f''(t)}{2}(t-c)^2dt">

Since

<img src="http://latex.codecogs.com/gif.latex?f%27%27%28x%29%20%5Cge%200" title="f''(x) \ge 0"> on the whole interval [0,1], it follows that the integral

lt;img src="http://latex.codecogs.com/gif.latex?%5Cint_c%5Ex%20%5Cfrac%7Bf%27%27%28t%29%7D%7B2%7D%28t-c%29%5E2dt%20%5Cge%200" title="\int_c^x \frac{f''(t)}{2}(t-c)^2dt \ge 0"> ,

hence

<img src="http://latex.codecogs.com/gif.latex?f%28c%29=f%28x%29+f%27%28c%29%28x-c%29%20%5Cge%200" title="f(c)=f(x)+f'(c)(x-c) \ge 0">

<img src="/cgi-bin/mimetex.cgi?f%28c%29=f%28x%29+f%27%28c%29%28x-c%29+%5Cint_c%5Ex%20%5Cfrac%7Bf%27%27%28t%29%7D%7B2%7D%28t-c%29%5E2dt" title="f(c)=f(x)+f'(c)(x-c)+\int_c^x \frac{f''(t)}{2}(t-c)^2dt">

Since

<img src="http://latex.codecogs.com/gif.latex?f%27%27%28x%29%20%5Cge%200" title="f''(x) \ge 0"> on the whole interval [0,1], it follows that the integral

lt;img src="http://latex.codecogs.com/gif.latex?%5Cint_c%5Ex%20%5Cfrac%7Bf%27%27%28t%29%7D%7B2%7D%28t-c%29%5E2dt%20%5Cge%200" title="\int_c^x \frac{f''(t)}{2}(t-c)^2dt \ge 0"> ,

hence

<img src="http://latex.codecogs.com/gif.latex?f%28c%29=f%28x%29+f%27%28c%29%28x-c%29%20%5Cge%200" title="f(c)=f(x)+f'(c)(x-c) \ge 0">

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