Derivatives& the Slope of the Graph: Inflection Point

1. Mar 29, 2010

MitsuShai

1. The problem statement, all variables and given/known data
For the function
f(x)=(x^2-3)/(x-2),
determine the locations of any points of inflection, if any

Last edited: Mar 29, 2010
2. Mar 29, 2010

crims0ned

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

Last edited: Mar 29, 2010
3. Mar 29, 2010

MitsuShai

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

4. Mar 29, 2010

crims0ned

$$f(x)=(x^2-3)(x-2)$$ is a polynomial and therefore it's domain should be all real number and this should carry on to it's derivatives as well.

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

5. Mar 29, 2010

MitsuShai

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

6. Mar 29, 2010

crims0ned

okay, so we have $$f(x)=\frac{(x^2-3)}{(x-2)} and f''(x)=\frac{2(x-2)}{(x-2)^4}$$ Even though x=2 is a vertical asymptote it is still a critical point on the second derivative

7. Mar 29, 2010

Staff: Mentor

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. Mar 29, 2010

MitsuShai

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. Mar 29, 2010

MitsuShai

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. Mar 29, 2010

crims0ned

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. Mar 29, 2010

MitsuShai

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. Mar 29, 2010

Staff: Mentor

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. Mar 29, 2010

crims0ned

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. Mar 29, 2010

MitsuShai

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

15. Mar 29, 2010

MitsuShai

This is impossible.

16. Mar 29, 2010

Staff: Mentor

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. Mar 29, 2010

MitsuShai

then what else can be the inflection point?

18. Mar 29, 2010

Staff: Mentor

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. Mar 29, 2010

MitsuShai

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

Last edited: Mar 29, 2010