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MitsuShai
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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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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?
[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.
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....
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)?
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
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?
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.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
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.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.
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.