# 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