Does the Function X + x^(2/3) Have a Concavity?

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What is the concavity of this function?

X+x2/3? I got the second derivative, and I. Did not gvet a value. Does this function not have concavity?

Note:2/3 is an exponent.
 
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Hi realism877! :smile:

realism877 said:
What is the concavity of this function?

X+x2/3? I got the second derivative, and I. Did not gvet a value. Does this function not have concavity?

Note:2/3 is an exponent.

Note the x2 and x2 buttons in the edit screen. They allow you to write subscripts/supscripts.

Anyway, you calculated the second derivative and you "don't get a value". What do you mean with that?? What is the second derivative of the function?

Do you mean that you don't find a point where the second derivative is 0? Well, that's certainly true.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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