prove: as the function increases then the graph is concave up/down

**tehbrosta** · October 5th 2008, 10:19 AM

how do I even start this one?

**bkarpuz** · October 5th 2008, 10:45 AM

Originally Posted by tehbrosta

how do I even start this one?

I will just give the solution for 2. (a) since the remaining case is very similar.
Clearly, if $f$ is increasing then we know that $f(x)\geq f(a)$ for all $x\in[a,b]$ .
Hence, we have
$\int\limits_{a}^{b}f(x)dx\geq\int\limits_{a}^{b}f( a)dx=f(a)(b-a),$
which completes the first part of the inequality.
For the remaining part, we consider that $f$ is concave down (convex).
This indicates that $f(x)\leq\frac{f(b)-f(a)}{b-a}(x-a)+f(a)$ holds for all $x\in[a,b]$ (draw a graph for the increasing convex function $f$ and the line passing at the points $(a,f(a))$ and $(b,f(b))$ ).
Integrating both sides of this inequality, we get
$\int\limits_{a}^{b}f(x)dx\leq\int\limits_{a}^{b}\B igg(\frac{f(b)-f(a)}{b-a}(x-a)+f(a)\Bigg)dx=(b-a)\frac{f(b)+f(a)}{2},$
which completes the proof.

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