Derivatives& the Slope of the Graph: Inflection Point

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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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Answers and Replies

  • #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.
 
  • #4
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[tex] f(x)=(x^2-3)(x-2) [/tex] is a polynomial and therefore it's domain should be all real number and this should carry on to it's derivatives as well.

[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) [/tex] is a polynomial and therefore it's domain should be all real number and this should carry on to it's derivatives as well.

[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)
 
  • #6
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okay, so we have [tex] f(x)=\frac{(x^2-3)}{(x-2)} and f''(x)=\frac{2(x-2)}{(x-2)^4} [/tex] Even though x=2 is a vertical asymptote it is still a critical point on the second derivative
 
  • #7
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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


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....
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)?
 
  • #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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okay, so we have [tex] f(x)=\frac{(x^2-3)}{(x-2)} and f''(x)=\frac{2(x-2)}{(x-2)^4} [/tex] Even though x=2 is a vertical asymptote it is still a critical point on the second derivative

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.
 
  • #10
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You're right it shouldn't be an inflection point because the original functions tangent line doesn't at x=2, but possibly it's asking for all critical points on the second derivative? Is this some kind of online homework program?
 
  • #11
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You're right it shouldn't be an inflection point because the original functions tangent line doesn't at x=2, but possibly it's asking for all critical points on the second derivative? Is this some kind of online homework program?

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
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okay, so we have [tex] f(x)=\frac{(x^2-3)}{(x-2)} and f''(x)=\frac{2(x-2)}{(x-2)^4} [/tex] Even though x=2 is a vertical asymptote it is still a critical point on the second derivative
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.
 
  • #13
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Isn't the definition of a critical point a point on f(x) that occurs at x0 if and only if either f '(x0) is zero or the derivative doesn't exist at that point?
 
  • #14
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Isn't the definition of a critical point a point on f(x) that occurs at x0 if and only if either f '(x0) is zero or the derivative doesn't exist at that point?

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
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Isn't the definition of a critical point a point on f(x) that occurs at x0 if and only if either f '(x0) is zero or the derivative doesn't exist at that point?
For x0 in the domain of f, x0 is a critical point if either f'(x0) = 0 or f' is not defined at xp.

The trouble is that x = 2 is not in the domain of the original function, f(x) = (x2 - 3)/(x - 2), so couldn't possibly be an inflection point.
 
  • #17
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For x0 in the domain of f, x0 is a critical point if either f'(x0) = 0 or f' is not defined at xp.

The trouble is that x = 2 is not in the domain of the original function, f(x) = (x2 - 3)/(x - 2), so couldn't possibly be an inflection point.

then what else can be the inflection point?
 
  • #18
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There isn't one. The only place the concavity changes is at x = 2, but x = 2 isn't in the domain of the original function.
 
  • #19
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There isn't one. The only place the concavity changes is at x = 2, but x = 2 isn't in the domain of the original function.

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