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## Homework Statement

For the function

f(x)=(x^2-3)/(x-2),

determine the locations of any points of inflection, if any

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- Thread starter MitsuShai
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- #1

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For the function

f(x)=(x^2-3)/(x-2),

determine the locations of any points of inflection, if any

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- #2

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Your function is [tex] f(x)=(x^2-3)(x-2) [/tex] right? Maybe if you foil it out first before you take the derivative?

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- #3

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Your function is [tex] f(x)=(x^2-3)(x-2) [/tex] right? Maybe if you foil it out first before you take the derivative?

The second derivative is right though; I entered it in and it was correct.

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[tex] f(x)=(x^2-3)(x-2) = x^3-2x^2-3x+6 [/tex] right? So there shouldn't be a quotient at all.

- #5

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[tex] f(x)=(x^2-3)(x-2) = x^3-2x^2-3x+6 [/tex] right? So there shouldn't be a quotient at all.

oh the original function is: f(x)=(x^2-3)/(x-2)

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- #7

Mark44

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Is there a typo in your functions? You have f(x)=(x^2-3)(x-2), but from what you have for f''(x), did you mean f(x) = (x^2 - 3)/(x -2)?## Homework Statement

For the function

f(x)=(x^2-3)(x-2),

determine the locations of any points of inflection, if any

## Homework Equations

f ''(x) = (2(x−2))/(x−2)^4

## The Attempt at a Solution

ok the concavity is (2, infinity) concave up

and (-infinity, 2) concave down, I entered them in as my answer and they were right

as for the inflection point, 2 is undefined, so I assumed that the function does not have an inflection point. I typed in 0 with a slash through it/ "{}" as my answer. My professor said that represents that there is no inflection point. But I got it wrong and I don't see another inflection point on the graph....

- #8

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Is there a typo in your functions? You have f(x)=(x^2-3)(x-2), but from what you have for f''(x), did you mean f(x) = (x^2 - 3)/(x -2)?

sorry, yes that was a typo

f(x)= (x^2−3)/(x-2)

f '(x)= [(x−3)(x−1)]/(x-2)^2

f ''(x) = (2(x−2))/(x−2)^4

the derivatives are all correct

- #9

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yes it is a critical point and it's where the function shifts from negative to positive, but it is an inflection point?

It's not continuous at x=2.

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- #11

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yes it is. This is the exact question (copied and pasted):

determine the locations of any points of inflection, if any.. Give your answer as a list of x values, e.g., {6,7,8,9}

- #12

Mark44

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No, that's not true. 2(x - 2)/(x - 2)^4 is identically equal to 2/(x - 2)^3. x = 2 is not in the domain of either function, so is not a critical number for the 2nd derivative. It's true that the concavity changes on either side of x = 2 for the original function, but for x = c to be an inflection point, f has to be defined at c. IOW, I agree with you that the function has no inflection point.

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yeah, I think 2 is a critical point because I used that to find concavity and I got the concavity questions right.

- #15

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This is impossible.

- #16

Mark44

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For x

The trouble is that x = 2 is not in the domain of the original function, f(x) = (x

- #17

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For x_{0}in the domain of f, x_{0}is a critical point if either f'(x_{0}) = 0 or f' is not defined at x_{p}.

The trouble is that x = 2 is not in the domain of the original function, f(x) = (x^{2}- 3)/(x - 2), so couldn't possibly be an inflection point.

then what else can be the inflection point?

- #18

Mark44

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- #19

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I e-mailed to my professor about it and we were right, just so you know :)

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