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

MATH110 April 2018

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Question 07 (e)
Let $f(x)={\frac {e^{x}}{x}}$ . (e) Given that $f''(x)={\frac {e^{x}(x^{2}-2x+2)}{x^{3}}},$ and the fact that $f''(x)\neq 0$ for all $x$ , find the intervals where $f$ is concave up and the intervals where it is concave down.

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Hint
If $f''(x)>0$ on an interval $(a,b)$ , then $f$ is concave up on the interval. On the other hand, if $f''(x)<0$ on an interval $(c,d)$ , then $f$ is concave down 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. Since $e^{x}>0$ and $x^{2}-2x+2=(x^{2}-2x+1)+1=(x-1)^{2}+1>0$ for any $x$ , the sign of the numerator in $f''(x)={\frac {e^{x}(x^{2}-2x+2)}{x^{3}}}$ is always positive. In other words, the sign of the second derivative is determined by that of the denominator. Since $x^{3}>0$ when $x>0$ and $x^{3}<0$ when $x<0$ , we have $f''(x)>0,\quad x>0,\quad {\text{and }}\quad f''(x)<0,\quad x<0.$ Namely, $f$ is concave down on $(-\infty ,0)$ , while it is concave up on $(0,\infty )$ . Answer: $\color {blue}{\mbox{CCD: }}(-\infty ,0),{\mbox{CCU: }}(0,\infty )$