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

MATH104 December 2014

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Question 05 (f)
Let $f(x)={\frac {e^{x}}{x^{2}}}$. Sketch the graph of $f(x)={\frac {e^{x}}{x^{2}}}$.

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
Combine all facts from $a$ to $e$ and see what they tell you about $f$.

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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 know from part $a$ that the critical point of $f$ occurs at $x=2$; hence the graph of the function will change from decreasing to the left of $x=2$ to increasing to the right of $x=2$ (see first image below). Next, we know that in the interval to the left of the vertical asymptote at $x=0$, the function is increasing. Moreover, in that interval, as $x$ tends to $-\infty$, $f(x)$ tends to $0$. From part $d$, we know that in this interval, the function is concave up (see second image below). Next, we focus on the intervals to the right of the vertical asymptote. In the interval $(0,2)$, we recall that the function is decreasing whilst in the interval $(2,\infty )$, the function is increasing (see third image below). From part $d$, we know that it is concave up in $(0,\infty )$, and flies off to infinity as $x$ tends to $\infty$. Combining these facts we have the graph depicted in the fourth image below: f(x) is decreasing to the left of x=2, and is increasing to the right of x=2 f(x) tends to 0 as x tends to negative infinity and is concave up; there is also a vertical asymptote at x=0 f(x) is increasing in (2, infinity) Final plot of f(x)

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