MHB Determine the location and nature of turning point

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SUMMARY

The turning point of the function \( (1 - \ln x)^2 \) for \( x > 2 \) occurs at \( x = e \) where \( \ln x = 1 \). This point is a minimum since the function is non-negative and equals zero at this point. To find the derivative, apply the chain rule: differentiate the outer function \( f(x) = x^2 \) to get \( 2x \), then substitute \( x \) with \( 1 - \ln x \) to obtain \( 2(1 - \ln x) \), and multiply by the derivative of \( 1 - \ln x \).

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives
  • Familiarity with the chain rule in differentiation
  • Knowledge of logarithmic functions and their properties
  • Basic algebra skills for solving equations
NEXT STEPS
  • Review the chain rule for derivatives in calculus
  • Study the properties of logarithmic functions, particularly \( \ln x \)
  • Practice finding turning points of functions using derivatives
  • Explore optimization problems in calculus to understand minima and maxima
USEFUL FOR

Students studying calculus, particularly those focusing on derivatives and optimization, as well as educators looking for examples of turning points in functions.

markosheehan
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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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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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