Find all extrema for the function $f(x)=-\frac{x^{3}}{3}+x^{2}-x+4$ Find all extrema for the function $f(x)=-\frac{x^{3}}{3}+x^{2}-x+4$ on the domain $x \in [-3.3]$.
Solution: $f'(x)=-x^{2}+2x-1 = 0 \implies (x-1)^{2}=0 \implies x^{*}=1$. 
Is that it?  
 A: Hint:
Next, I think you need to show that this isn't merely an infection point, such as (0,0) on $y=x^3$  
To do this, try taking the second derivative of $f'(x)=-x^{2}+2x-1$  
$$f''(x)=-2x+2$$  
then substitute $x:=1$ to test the point you discovered. Is it a relative maximum, a minimum or an inflection point?  
$$-2(1)+2=0$$  
That the result is $0$ shows that the slope is neither decreasing nor increasing at this point. At a relative maximum or minimum you would have a non-zero result. So you are dealing with an inflection point. Since this is the only critical point (as identified by the first derivative), the function is monotonically increasing or decreasing throughout the interval. Calculate the values for $f(x)$ at the extrema, and you have your max/mins. (Since $f(-3)>f(3)$, you know the function is decreasing.)
The following diagram bears this out. The inflection point is marked. The function can be seen to be monotonically decreasing. 

A: Hint:
note that $f'(x)=-(x-1)^2 \le 0 \quad \forall x \in \mathbb{R}$, so the function is motononic decreasing and $x=1$ can not be an extremum.
You have to find the values of the function in $x=-3$ and $x=3$ to find the extrema in the given interval.
