Functions with asymptotes and differentiability Question

In summary, In 2b, there are three separate parts to the graph: the first is concave upwards to the left of -1, the second 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 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.
  • #1
ardentmed
158
0
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:
08b1167bae0c33982682_9.jpg


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.
 
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  • #2
Does anyone know how to go about doing this question?

Thanks in advance.
 
  • #3
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.
 
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  • #4
Rido12 said:
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.
 
  • #5
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.
 
  • #6
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
 

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  • #7
Correct, again.
 
  • #8
Rido12 said:
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
 

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  • #9
Great job! That looks right. (Nod)
 

FAQ: Functions with asymptotes and differentiability Question

What is an asymptote?

An asymptote is a line that a graph approaches but never touches. It can be vertical, horizontal, or slanted.

How do you determine the equation of an asymptote?

The equation of a vertical asymptote can be found by setting the denominator of the function equal to zero and solving for the variable. The equation of a horizontal asymptote can be found by taking the limit of the function as x approaches positive or negative infinity.

What is differentiability?

Differentiability refers to the smoothness of a function. A function is differentiable if it has a well-defined derivative at every point in its domain.

How do you determine if a function is differentiable?

A function is differentiable if it is continuous and has a well-defined derivative at every point in its domain. This can be checked by finding the derivative of the function and ensuring that it exists at all points in the domain.

Can a function have an asymptote and still be differentiable?

Yes, a function can have an asymptote and still be differentiable. Asymptotes only describe the behavior of the function as x approaches infinity or a specific value, while differentiability looks at the behavior of the function at every point in its domain.

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