(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

f is a continues function in [a,b] and has a second derivative in (a,b). L(x) is a line that goes through (a,f(a)) and (b,f(b)).

Prove that if f''(x)>0 in (a,b) then L(x)>f(x) for every x in (a,b)

2. Relevant equations

MVT

3. The attempt at a solution

First of all,

[tex] L(x) = f(a) + \frac{f(b)-f(a)}{b-a} (x-a) [/tex]

and so

[tex] L(x) - f(x) = f(a) - f(x) + \frac{f(b)-f(a)}{b-a} (x-a) [/tex]

[tex] \frac{f(b)-f(a)}{b-a} = f'(t) [/tex] where t is in (a,b)

and so

[tex] ( L - f)' (x) = f'(t) - f'(x) [/tex]

Now, [tex] ( L - f )' (x) = 0 [/tex] only when [tex] f'(t) = f'(x) [/tex]

And since f''(x)>0 f'(x) is an injective function in (a,b) and so [tex] f'(t) = f'(x) [/tex] only when x=t. and since f''(x)>0 we get a maximum at x=t.

Now, (L-f)(a) = 0 and (L-f)(b) = 0. If for any other x_0 =/= t

(L-f)(x_0)=0 then that would mean that for some x in (a,x_0) and for some x in (x_0,b) (L-f)'(x) = 0, but this is impossible as (L-f)'(x) is injective and we alredy found one point (t) where (L-f)'(x) = 0.

Also, (L-f)(t) > (L-f)(a) = 0 because it's a maximum in (a,b). And so for all x in (a,b) L-f)(x) > 0 => L(x) > f(x).

I felt that that proof was pretty weak, especially towards the end. How can I make it better. And is there any easier way?

Thanks.

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