MHB Determine the location and nature of turning point

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The turning point of the function (1 - ln x)² occurs when ln x = 1, which corresponds to x = e. The derivative is computed using the chain rule, where the outer function is f(x) = x², leading to 2(1 - ln x) as part of the derivative. This derivative must then be multiplied by the derivative of the inner function, 1 - ln x. The analysis shows that for x > 2, the point x = e represents a minimum. Understanding these concepts is crucial for solving calculus problems involving turning points.
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whats the turning point of (1-lnx)², x>2 is it a minimum or maximum. can someone help me
 
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Re: calculus

We have $$(1-\ln x)^2\ge0$$ and, moreover, $(1-\ln x)^2=0\iff \ln x=1$. Can you find the solution of $\ln x =1$?
 
how do you get the derivative of it
 
As for any composition of functions. The outer function if $f(x)=x^2$, so first you differentiate $x^2$, and the result is $2x$. Now our whole function is not $f(x)$, but $f(1-\ln x)$. Therefore you replace $x$ in $2x$ with $1-\ln x$ to get $2(1-\ln x)$. To finish, you need to multiply this by the derivative of $1-\ln x$.

I recommend reviewing the chain rule for computing the derivative of the composition of functions in your textbook.
 

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