Maximizing a Continuous Function on A: Analysis Homework Help

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Homework Statement



Let A= [0,1). Find a continuous function g: from A to R that does not attain a maximum value.

Homework Equations





The Attempt at a Solution


I believe that g(x)=x or g(x)=e^x represent such a function, but I do not know how to use the IVT to prove that either of them work. Please help! I need as much information as possible without completely giving away the answer.
 
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Hi Janez25! Welcome to PF! :wink:
Janez25 said:
… I believe that g(x)=x or g(x)=e^x represent such a function, but I do not know how to use the IVT to prove that either of them work.

Yes, either will do :smile:

now, what property (of either) means that the maximum is never reached?

(i don't think you need the IVT for this)
 


I am not sure what property I should use. I also am not sure how to show that either function does not attain a maximum value on the interval [0,1) formally.
 
Janez25 said:
I am not sure what property I should use. I also am not sure how to show that either function does not attain a maximum value on the interval [0,1) formally.

Hint: asssume it does attain its maximum, or any local maximum, at x = a in [0,1) …

is that possible for x or for ex?

what property would f(x) have if that isn't possible?
 
Thanks for your help!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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