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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.

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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 at 3:44
    
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 $g(x) = x - 1/2$. –  Bungo Aug 22 at 6:05

1 Answer 1

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.

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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 at 3:48
    
@user131785 I think it is... There are many functions that are correct (e.g. $g(x)=\dfrac{x^2+2x+1}{x+1}$) –  JChau Feb 27 at 3:52
    
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 at 3:55
    
@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. –  JChau Feb 27 at 3:58
    
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 at 4:06

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