Algotssleeddynf

2022-01-05

I cannot prove that if $f\left(x\right)$ is convex on $[a,b]$ then

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int}_{a}^{b}f\left(x\right)dx\le \frac{f\left(a\right)+f\left(b\right)}{2}$

otoplilp1

Beginner2022-01-06Added 41 answers

The inequality

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int}_{a}^{b}f\left(x\right)dx$

is a special case of Jensens

is a special case of Jensens

sukljama2

Beginner2022-01-07Added 32 answers

For convenience, we can take $b=1,\text{}a=-1$

For the first inequality, use

$f\left(0\right)\le \frac{f\left(x\right)+f(-x)}{2}$

and integrate both sides from$x=-1$ to 1

For the second, use

$f\left(x\right)=f(\frac{1-x}{2}(-1)+\frac{1+x}{2}\left(1\right))\le \frac{1-x}{2}f(-1)+\frac{1+x}{2}f\left(1\right)$

and integrate both sides from$x=-1$ to 1.

For the first inequality, use

and integrate both sides from

For the second, use

and integrate both sides from

star233

Skilled2022-01-11Added 403 answers

Since f is convex ob [a,b], one has:

Therefore, one gets:

Finally, one has:

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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