MHB Functions with asymptotes and differentiability Question

[ardentmed]

Hey guys,

More questions for you guys this time, these seem easy but always have a few nuances I seem to miss. With that said, I'd greatly appreciate your guys' help.

Question:

For 1a, I sketched two straight lines where x=/ 1 and y=1/2 for one line and y= -1/2 for the other.

Thus, x=1 is a vertical asymptote.

For 1b, f is not continuous because the limit at 1 from the left and right differ; one is 1/2 while the other is -1/2.

For 1c, the function is not differentiable at a=1 because a function f is differentiable at "a" if f ' (a) exists, meaning it must be differentiable over an open interval with "a" in it. a=1 is thus not part of the domain of f: (-infinity,1)u(1, infinity). Thus, f ' (1) is not possible.
As for 2a, I took the limit as x approaches infinity and used the limit division laws (I took the limit of both the nominator and the denominator accordingly)

Thus, $\lim_{{x}\to{infinity}}$ (1+ 2/x) / $\lim_{{x}\to{infinity}}$ √(9+ 5/(x^2)) =
1/√9
=1/3


Also, 7b gave me the most bizarre graph I've seen in quite a while. Would anyone know what this should roughly look like? The one I sketched has 3 points of discontinuity, which I find disconcerting.


Thanks in advance.

 

[ardentmed]

Does anyone know how to go about doing this question?

Thanks in advance.

 

[Dethrone]

1a) You are correct about the graph, but you have to specify the interval of which y = 1/2 and y = -1/2, which I'm sure you probably did, but didn't bother typing out.

1b) and c) and 2a are also correct.
2b) is quite weird, actually. If I were to explain what I drew, I would be basically repeating what the question specifies, but I will try. There's three separate parts to the graph. The first part is concave upwards to the left of -1, the second part is a line that goes from y = 5 (closed dot, since it's continuous from the right.) down to negative infinity when x = 2. And the last portion is a line that goes from positive infinity to -2. There is a vertical asymptote at x = -1 and 2. Also a horizontal asymptote of y = 2 when you approach to negative infinity, and y = -2 when you approach to positive infinity.

 

[ardentmed]

1a) You are correct about the graph, but you have to specify the interval of which y = 1/2 and y = -1/2, which I'm sure you probably did, but didn't bother typing out.

1b) and c) and 2a are also correct.
2b) is quite weird, actually. If I were to explain what I drew, I would be basically repeating what the question specifies, but I will try. There's three separate parts to the graph. The first part is concave upwards to the left of -1, the second part is a line that goes from y = 5 (closed dot, since it's continuous from the right.) down to negative infinity when x = 2. And the last portion is a line that goes from positive infinity to -2. There is a vertical asymptote at x = -1 and 2. Also a horizontal asymptote of y = 2 when you approach to negative infinity, and y = -2 when you approach to positive infinity.

But isn't 1a a hole, not a vertical asymptote, since it cancels out in the expression?

Thanks again for the help.

 

[Dethrone]

I didn't say it was a vertical asymptote. I just said that because the function is an absolute value, it's defined as a piecewise function. It takes on a different function depending on which region of the domain it lies on, which you didn't specify.

 

[ardentmed]

Here is the (inverted) graph for this function. Sorry about the webcam quality. Am I on the right track?

Thanks so much for the help. View attachment 2911

 

[Dethrone]

Correct, again.

 

[ardentmed]

1a) You are correct about the graph, but you have to specify the interval of which y = 1/2 and y = -1/2, which I'm sure you probably did, but didn't bother typing out.

1b) and c) and 2a are also correct.
2b) is quite weird, actually. If I were to explain what I drew, I would be basically repeating what the question specifies, but I will try. There's three separate parts to the graph. The first part is concave upwards to the left of -1, the second part is a line that goes from y = 5 (closed dot, since it's continuous from the right.) down to negative infinity when x = 2. And the last portion is a line that goes from positive infinity to -2. There is a vertical asymptote at x = -1 and 2. Also a horizontal asymptote of y = 2 when you approach to negative infinity, and y = -2 when you approach to positive infinity.

And here is the graph for 2b. View attachment 2912

 

[Dethrone]

Great job! That looks right. (Nod)