Concavity and inflection points for f(x)=x-lnx

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Homework Statement


determine the intervals on which the function is concave up or down and find the points of inflection.
f(x)= x-lnx for x>0


The Attempt at a Solution



f '(x)=1 -(1/x)
f"(x)= 1/x^2

I know to find concavity and inflection points you take the second derivative and set it equal to 0, then find intervals w/ the inflection point. But I don't know if I just took the derivative wrong but right now I can't solve the second derivative...any help, please?
 
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Think about what the limiting values are for each of the terms as x goes to 0, infinity, 1, etc. Then try to sketch a plot of the curve, and plot it using a calculator to see how close you were. You'll be forced to think about how the function behaves and it might show you more than playing around with the algebra.
 
I get what you're saying, but I really need to know how to solve it with algebra...
 
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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