Science:Math Exam Resources/Courses/MATH100 A/December 2023/Question 27(a)

MATH100 A December 2023

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[hide] Question 27(a)
Let $f(x)=8xe^{-x/2},\qquad x\geq 0.$ (This is a scaled example of the gamma distribution from probability.) In this question, you may use without proof the fact that $f'(x)=-4e^{-x/2}(x-2).$ (a) List the vertical and horizontal asymptotes. Leave the field blank if there is no such asymptote. Note: For part (a), you may simply list your answers without justification.

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!

[show] Hint

Recall the definitions. A function $f$ has a vertical asymptote at $x=c$ if $\displaystyle \lim _{x\rightarrow c^{-}}f(x)=\pm \infty$ or $\displaystyle \lim _{x\rightarrow c^{+}}f(x)=\pm \infty$. What does this mean for $f(x)$ if $x$ is near c? Can $f$ be continuous at $c$? A function $f$ has a horizontal asymptote as $x\rightarrow -\infty$ or $x\rightarrow \infty$ if the limit $\displaystyle \lim _{x\rightarrow -\infty }f(x)$ exists or if the limit $\lim _{x\rightarrow \infty }f(x)$ exists, respectively. Can you try to compute these limits?

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[show] 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 the domain of $f$ is the non-negative real numbers, the function can only have a horizontal asymptote at $\infty$, so there is no need to consider the limit as $x$ approaches $-\infty$. To compute the limit, we may rewrite $f(x)$ as $f(x)={\frac {8x}{e^{x/2}}}.$ You may know that the above limit evaluates to 0, if you understand that the exponential function grows much more rapidly than any polynomial. Alternatively, we may use L'Hôpital's rule to compute $\lim _{x\rightarrow \infty }f(x)=\lim _{x\rightarrow \infty }{\frac {8}{e^{x/2}\cdot {\frac {1}{2}}}}=0.$ Thus, our function has the horizontal asymptote $\{y=0\}$ at $\infty$. Since the function is continuous on its domain, it has no vertical asymptotes.