Science:Math Exam Resources/Courses/MATH110/April 2018/Question 07 (c)

MATH110 April 2018

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Question 07 (c)
Let $f(x)={\frac {e^{x}}{x}}$. (c) Find the intervals where $f$ is increasing and the intervals where it is decreasing.

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 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
If $f'(x)>0$ on an interval $(a,b)$, then $f$ is increasing on the interval. On the other hand, if $f'(x)<0$ on an interval $(c,d)$, then $f$ is decreasing on the interval.

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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 first find the derivative of $f$. Using the quotient rule, we get $$f'(x)={\frac {(e^{x})'\cdot x-e^{x}\cdot (x)'}{x^{2}}}={\frac {xe^{x}-e^{x}}{x^{2}}}={\frac {(x-1)e^{x}}{x^{2}}}.$$ Since $e^{x}$ is always positive and $x^{2}$ is positive except for $x=0$, the sign of the first derivative depends on $x-1$. Indeed, $f'(x)>0$ if $x>1$ while $f'(x)<0$ if $x<0$ or $0<x<1$. Therefore, the statement in the Hint implies that $f$ is increasing on $(1,\infty )$ and decreasing on $(-\infty ,0)\cup (0,1)$. Note that $f$ is undefined at $x=0$, because the denominator vanishes at this point. Answer: $\color {blue}{\mbox{INC: }}(1,\infty ),{\mbox{DEC: }}(-\infty ,0)\cup (0,1)$