Science:Math Exam Resources/Courses/MATH100/December 2014/Question 08 (e)

MATH100 December 2014

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Question 08 (e)
Let $f(x)={\frac {\log(x)}{\sqrt {x}}}$. Note $\log(x)$ is the natural logarithmic function, also denoted $\ln(x)$ (e) Find the intervals where $f(x)$ is increasing, and the intervals where $f(x)$ is decreasing. Find the coordinates of local maximum and minimum if they exist.

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Consider the derivative of $f(x)$ and how it being positive or negative relates to $f(x)$ being increasing or decreasing.

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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. First find the derivative ${\begin{aligned}f'(x)&={\frac {{\sqrt {x}}\cdot {\frac {1}{x}}-{\frac {1}{2{\sqrt {x}}}}\log(x)}{x}}\\&={\frac {2-\log(x)}{2x^{3/2}}}\end{aligned}}$ So the derivative exists everywhere on the domain. Any local maximum or minimum occur when $f'(x)=0$. This is zero when $2-\log(x)=0$, that is, when $x=e^{2}$. ${\begin{aligned}(0,e^{2}):f'(x)&={\frac {pos}{pos}}=pos&{\text{increasing}}\\(e^{2},\infty ):f'(x)&={\frac {neg}{pos}}=neg&{\text{decreasing}}\end{aligned}}$ Hence $x=e^{2}$ gives the only local maximum at the coordinates $(e^{2},f(e^{2}))=(e^{2},{\frac {2}{e}})$ and there is no local minimum.

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