# Thread: multiple choice on stationary points

1. ## multiple choice on stationary points

x^3 -6x^2 +12x +18
Which one of these choices are correct( Choose A,B,C OR D only)
A. X=2 is a point of inflection
B. X=2 is an upward turning point and x=0 is a downward turning point
C. X=2 is a downward turning point and x=0 is a upward turning point
D. X=2 can not be characterised because f’’(x)=0

The graph are below: Showing the point at X=2

Thanks a lot for answering the question.

2. Originally Posted by firebrend
x^3 -6x^2 +12x +18
Which one of these choices are correct( Choose A,B,C OR D only)
A. X=2 is a point of inflection
B. X=2 is an upward turning point and x=0 is a downward turning point
C. X=2 is a downward turning point and x=0 is a upward turning point
D. X=2 can not be characterised because f’’(x)=0

The graph are below: Showing the point at X=2

3. my answer is A, because at x=2 f'(x)=0 and f''(x)=0==> the point is stationary point but is neither maximum nor minimum point. therefore it is an inflection point. What about your answer? do you agree with my reasoning? thanks

4. Originally Posted by firebrend
my answer is A, because at x=2 f'(x)=0 and f''(x)=0==> the point is stationary point but is neither maximum nor minimum point. therefore it is an inflection point. What about your answer? do you agree with my reasoning? thanks
I agree, to a point.

I still believe that it is necessary to show that f''(x) changes sign at x = 2 to confirm a point of inflection.

I can't say for sure that there is not some function that exists where f'(a) = 0, f''(a) = 0 , has no extrema at (a,f(a)) , and that does not have an inflection point at x = a ... I just can't think of a rigorous proof of your statement or a suitable counterexample.

5. If $\displaystyle f(x)= x^3 -6x^2 +12x +18$ then $\displaystyle f'(x)= 3x^2- 12x+ 12= 3(x^2- 4x+ 3)= 3(x- 3)(x- 1)$ which is ot 0 at x= 2 so 2 is not a critical point and is not 0 at x= 0 so 0 is not a turning point. $\displaystyle f''(x)= 6x- 12= 6(x- 2)$. Yes, f''(2)= 0 and f''(x) changes sign at x= 2 (f''(x) is negative for x< 2, positive for x> 2).