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I am having some trouble figuring out a few math problems from my Calc 1 class. I am not sure where to start, as all the limits are different. enter image description here

  1. find a function that satisfies the given conditions and then sketch it.

  2. sketch a graph of the function y=f(x) that satisfies the given conditions. Just label the coordinate axes and sketch the appropriate graph.

For 60, 62, and 64 they are kinda the same thing. enter image description here 60: would it be a function where if you let X=-1, and the denominator =0, is that what we are looking for? like $ \frac{2}{x+1} \ . $

  1. I would say yes, even though g(x) and f(x) are not discontinuous on their own, that changes when you put them in a function together such as $ f(x)/g(x).$

  2. not to sure

Thank you for all your help.

user68061
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user131785
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  • For 60; would 2x/x+1 be an answer since it has a non removable discontinuity at x=-1 ? – user131785 Feb 27 '14 at 03:44
  • Simple example for 62: $f(x) = 1$, $g(x) = x$. Or, if you want the discontinuity in the interior of $[0,1]$, you can use $g(x) = x - 1/2$. –  Aug 22 '14 at 06:05

1 Answers1

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Just try drawing what the graph could potentially be on paper, and find an equation for it. Here's a freebie for the first one (number 74): $$g(x)=\dfrac{1}{x-3}$$ Note: I am not taking calculus yet, but I do know about limits and such... I don't know if there is a faster way because my way seems inefficient and not algebraic-like.

EDIT: For the second one, maybe this would work: $$f(x)=\begin{cases} -\frac{1}{x^2+\frac{1}{2}}, \ \ \ \ x < 0 \\ 0, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=0 \\ \frac{1}{x^2+\frac{1}{2}}, \ \ \ \ \ \ \ \ x > 0 \\ \end{cases}$$ YET ANOTHER EDIT: Your answer for the final one (nonremovable discontinuity) is correct.

Anonymous Computer
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  • so kinda guess and check? I will try for the second one. was the answer for 60 correct, 2x/x+1 correct? – user131785 Feb 27 '14 at 03:48
  • @user131785 I think it is... There are many functions that are correct (e.g. $g(x)=\dfrac{x^2+2x+1}{x+1}$) – Anonymous Computer Feb 27 '14 at 03:52
  • and that is discontinuous at x=-1 since the denominator would =0 there and we know the discontinuous is not removable since the top and bottom do not cancel fully? – user131785 Feb 27 '14 at 03:55
  • @user131785 To be totally honest with you I don't really know what a nonremovable discontinuity is. I think your answer $\dfrac{2x}{x-1}$ is better because it does not have any factors that can cancel out. I don't know if I am correct, but because my answer has factors that can cancel, it is removable. In any case you are right. – Anonymous Computer Feb 27 '14 at 03:58
  • ah now 1/x-3 was for 74, would would the same 1/x-2 work for 70? seems like it should – user131785 Feb 27 '14 at 04:06
  • @user131785 I really don't think so because first of all, the graph does not approach $-2$ when it gets to a negative number very close to $0$ like $-0.000001$. See my edited answer, maybe that is a solution? – Anonymous Computer Feb 27 '14 at 04:48
  • ah thank you, that looks pretty good, hopefully itll work – user131785 Feb 27 '14 at 21:27
  • that actually cant work, since if you plug 0 in you do not get 0 for x=0 – user131785 Feb 28 '14 at 00:44
  • @user131785 The second part of the function says that if $x=0$, $f(x)=0$. That means that $f(0)=0$, which meets the requirement for the function. If you still do not believe me, click here to go to Desmos Graphing Calculator, type a curly brace { in one of the boxes, and enter this piece of text inside the curly braces: x<0:-1/(x^2+0.5),x=0:0,x>0:1/(x^2+0.5). The graph should show, and you will see that $(0,0)$ is a point on the graph. – Anonymous Computer Feb 28 '14 at 01:46
  • but how about for the - and + infinity? – user131785 Feb 28 '14 at 03:07
  • @user131785 Just plot the graph at the website I told you about and you should see that $f(x)$ approaches $0$ when $x \to \pm\infty$ – Anonymous Computer Feb 28 '14 at 04:26