Science:Math Exam Resources/Courses/MATH104/December 2014/Question 05 (d)

MATH104 December 2014

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Question 05 (d)
Let $f(x)={\frac {e^{x}}{x^{2}}}$. Find the intervals on which $f$ is concave upward.

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
Find the intervals on which the second derivative $f''$ exists and is nonzero.

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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. Recall that when $f''>0$, $f$ is concave up; likewise, when $f''<0$, $f$ is concave down. So what we need to do is find the intervals on which $f''$ exists and is non-zero. From above, we know that $f''(x)={\frac {e^{x}(x^{2}-4x+6)}{x^{4}}}$, and indeed, $f''(x)=0$ if and only if $e^{x}=0$ or $x^{2}-4x+6=0$. Since $e^{x}>0$ for all $x$, we have that $x^{2}-4x+6=0$. Using the quadratic formula, we find that this quadratic equation has no real roots. Therefore, there is no $x$ where $f''(x)=0$. Now, we make a table as before for $f'$, noting that $f''$ does not exist at $x=0$. This gives us two intervals, namely $(-\infty ,0)$ and $(0,\infty )$ where $f''$ is nonzero and defined: $(-\infty ,0)$ $(0,\infty )$ $e^{x}$ + + $x^{2}-4x+6$ + + $x^{4}$ + + $f''$ + + $f$ Concave Up Concave Up So on both $(-\infty ,0)$ and $(0,\infty )$, $f$ is concave up.

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