Differentiability

Kristina123

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Are my answers sufficient to appropriately answer the question? Or, do I need to add more explanations? I want to explain further why it is differentiable and not differentiable

Thanks!
 
Are my answers sufficient to appropriately answer
Hi Kristina. Part (b) asks where function f is defined, yet you've answered where it's not defined. I would fix that.

For the second question, are you thinking that the function might not be differentiable at some values of x in the domain?

I want to explain further why it is differentiable and not differentiable
Why? The exercise does not ask for that.

?
 
Hi Kristina. Part (b) asks where function f is defined, yet you've answered where it's not defined. I would fix that.

For the second question, are you thinking that the function might not be differentiable at some values of x in the domain?


Why? The exercise does not ask for that.

?
I'm thinking that the function f is not differentiable when it approaches closer and closer to -2 and 2 because it seems as if the line is literally a vertical line, but when I graph the derivative of the function it appears that it is differentiable for the all x values of the domain in the function f. So does this mean that the function f doesn't have any areas where it is not differentiable?
 
I'm thinking that the function f is not differentiable when it approaches closer and closer to -2 and 2 because it seems as if the line is literally a vertical line
Thank you for your thoughts, Kristina. As long as x is still approaching -2 from the left or approaching +2 from the right, the function is continuous and differentiable. It doesn't matter how close we are to either asymptote (x=-2 or x=2). The function is not differentiable when x is exactly -2 or 2 (and, as you've indicated) at any value in between.

Therefore, I would cut short your posted answer for the second question, by writing just the first part: f(x) is differentiable everywhere on its domain.

Please post your corrected domain statement, so we can check it.

PS: It's not the graph that becomes vertical near -2 and 2. The asymptotes are the only vertical lines. The graph itself is never vertical, even when function values become infinitely large (in absolute value).

?
 
Thank you for your thoughts, Kristina. As long as x is still approaching -2 from the left or approaching +2 from the right, the function is continuous and differentiable. It doesn't matter how close we are to either asymptote (x=-2 or x=2). The function is not differentiable when x is exactly -2 or 2 (and, as you've indicated) at any value in between.

Therefore, I would cut short your posted answer for the second question, by writing just the first part: f(x) is differentiable everywhere on its domain.

Please post your corrected domain statement, so we can check it.

PS: It's not the graph that becomes vertical near -2 and 2. The asymptotes are the only vertical lines. The graph itself is never vertical, even when function values become infinitely large (in absolute value).

?
Thank you so much for your insightful response. My updated answer for b) is now:

1634505243434.png
 
the interval [-2,2] in between is undefined
I understand what you mean, but, technically speaking, the interval [-2,2] is always defined; it's the function that is not defined.

;)
 
I'm thinking that the function f is not differentiable when it approaches closer and closer to -2 and 2 because it seems as if the line is literally a vertical line, but when I graph the derivative of the function it appears that it is differentiable for the all x values of the domain in the function f. So does this mean that the function f doesn't have any areas where it is not differentiable?
Is the function defined when:

|| x|| < 2 .,..?
 
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