Is there an inflection point in this function at $x=1$? I am working on a calculus packet, and one of the questions is as follows:
The derivative of a polynomial function, $f(x)$, is represented by the equation $f'(x) = -2x(x-3)^2$.
Additionally, $f(1) = –3$ and the graph of $f(x)$ is concave up at $x = 1$.  At what value(s) of $x$ does the graph of $f(x)$ have a relative maximum? A relative minimum?
So, I used both the first derivative test and the 2nd derivative test to find that there is a maximum at $x=0$ and no minimums.  I even solved for $f(x)$ and graphed it to confirm my findings.
What I am confused by is why, in the original problem, it says "the graph of $f(x)$ is concave up at $x=1$."  It seems to me that $f(x)$ has an inflection point at $x=1$, and so therefore cannot be concave up at this point.  
Can anybody confirm this for me?  
 A: f'(x)=-2x(x-3)^2
f'(x)=-2x(x^2-6x+9)
f'(x)=-2x^3+12x^2-18x
f''(x)=-6x^2+24x-18
There is an inflection at the zero of the second derivative, -6x^2+24x-18
-6x^2+24x-18=0
A=-6 B=24 C=-18
Using the quadratic equation, inflection points occur at the coordinates (1,-3) and (3,-11)
The function is concave on the interval
(-∞,1)U(3,∞)
The function is convex on the interval (1,3)
So yes, you are indeed correct. I love when the student is right and the teacher is wrong.
Oh, and by the way, the relative maximum occurs at the coordinate (0,2.5) and there are no relative minimums since the only other extremum (point where the derivative is zero) is at the inflection point (3,-11).
Proof:
f'(x)=-2x(x-3)^2
-2x(x-3)^2=0
x= 0 or 3, but 3 is a double root
A: You are correct. It's a sufficient condition that the 2nd derivative of f(x) = 0 and that in this point(s) f(x-£) and f(x+£) when $£\rightarrow 0$ have different signs. I hate when they do that in textbooks!
A: The second derivative is
$$
f''(x)=-2(x-3)^2-4x(x-3)=-6(x-3)(x-1)
$$
The third derivative is
$$
f'''(x)=-12(x-2)
$$
Since $f''(0)=-6<0$, the function has a maximum at $0$; since $f''(3)=0$ and $f'''(3)\ne0$, $3$ is an inflection point; the same for $1$.
