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

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SUMMARY

The function f(x) = x - ln(x) for x > 0 exhibits specific concavity characteristics determined by its second derivative, f''(x) = 1/x². The function is concave up for all x > 0 since the second derivative is always positive in this interval. There are no points of inflection for this function as the second derivative does not equal zero for any x > 0. Understanding the behavior of this function requires analyzing its derivatives and considering limiting values as x approaches 0 and infinity.

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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...
 

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