Science:Math Exam Resources/Courses/MATH104/December 2010/Question 01 (i)

MATH104 December 2010

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Question 01 (i)
If the derivative of $f(x)$ is given by $f'(x)={\frac {2\ln(x)}{x}},$ find the interval or intervals on which $f(x)$ is concave down.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
Find the second derivative of f(x).

Hint 2
What do the signs of the second derivative tell you about concavity?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We have that $f'(x)={\frac {2\ln {}x}{x}}.$ The second derivative is (using quotient rule), $f''(x)={\frac {{\frac {2}{x}}x-2\ln {}x(1)}{x^{2}}}=2{\frac {1-\ln {x}}{x^{2}}}.$ In order to find intervals of concavity we must check where the second derivative is zero or does not exist. We see that the second derivative is zero whenever 1-lnx=0. This occurs if x=e. The second derivative doesn't exist if x=0 (the denominator vanishes) or if x<0 since then the logarithm cannot be evaluated. In fact we anticipate that x<0 is not even in the domain of the function. Therefore our intervals of interest are $(0,e),(e,\infty )$. We can make a table (or number line) as follows $\displaystyle {}(0,e)$ $(e,\infty )$ $1-\ln {x}$ $\displaystyle {}+$ $\displaystyle {}-$ $x^{2}$ $\displaystyle {}+$ $\displaystyle {}+$ $\displaystyle {}f''(x)$ $\displaystyle {}+$ $\displaystyle {}-$ $\displaystyle {}f(x)$ Concave Up Concave Down Therefore we have that f(x) is concave up on $\displaystyle {}(0,e)$ and concave down on $(e,\infty )$.

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