Finding extreme values where second derivative is zero Consider this function:
$$f(x)= 5x^6 - 18x^5 + 15x^4 - 10$$
I am told to find the extreme values of this function. So at first, I take the first derivative and set it zero. 
$$f'(x)=30x^5-90x^4+60x^3=0(say)$$
$$=x^5-3x^4+2x^3=0$$
If I evaluate for $x$, I find $x= 0, 1, 2$. Now let's take the second derivative of $f(x)$:
$$f''(x)= 150x^4-360x^3+180x^2$$
I am Ok with $f''(1)$ and $f''(2)$ because they give a negative and positive value respectively. Now, $f''(0) = 0$. So I don't know if I have a maxima or minima at $x=0$. I have found one way to solve this on internet that is named as first derivative test. Here I take two values of $x$, one $<0$, other $>0$.  Let me take $0.5$ and $-1$. 
$$f'(-1)= -180$$
$$f'(0.5)=45/16$$
So if $x<0$ then the function is decreasing. But when $0 < x <1$ (less than one because we have another maxima at $x=1$, but I am worrying only about that at $x=0$), it's increasing. Thus I can say we have a minima at $x=0$.
The main reason behind my asking is, after finding $f''(x)=0$, my solution book has shown that $f^{(iii)}(0)=0$ but $f^{(iv)}(0)= 360>0$. Thus we have a minima at $x=0$. But this is the INFLECTION POINT test, isn't it? But a point being an inflection point doesn't necessarily mean that it is a local maxima or local minima. So I would like to know if my work is right as I am not deeply familiar with the solution that I gave above and if my book's solution is also allowable or not?
And one more thing I am not sure about: Can a point be an inflection point and at the same time a local maxima or minima? If yes, an example would be very helpful.
 A: It has a local minimum at $0$, because $f^{(3)}(0)=0$ and $f^{(4)}(0)>0$.
The general rule here is: you keep derivating until you get the first $n$ such that $f^{(n)}(a)\neq0$. Then:


*

*if $n$ is odd, $a$ is neither a local maximum nor a local minimum;

*if $n$ even and $f^{(n)}(a)>0$, then $f$ has a local minimum at $a$;

*if $n$ even and $f^{(n)}(a)<0$, then $f$ has a local maximum at $a$.

A: Close to zero, discarding the smaller terms, the function is essentially 
$$15x^4-10,$$ with derivatives $60x^3$ and $180x^2$.
Clearly, the second derivative remains positive, the first derivative is increasing and changes sign (from negative to positive) and the function itself has a minimum, as it is decreasing then increasing.
The plot shows you the true function, the approximation at $x=0$ and its derivatives (I let you guess which is which).

A: Consider $f(x)$ whose first n derivations are zero in $x_0$. An even n implies that point is inflection point and an odd n implies it to be maximum or minimum. then to prove which case is happening exactly you need to calculate $f^{n+1}(x)$. Positivity of it implies minimum and negativity implies maximum
A: For a point to be inflection point you need to have f'' has different signs on two sides of that point. Merely showing that f''(x)=0 does not guarantee an inflection point.
For your final question, it's a duplicate of Is it possible to be both a relative max/min and an inflection point?, check that post for a constructed function with inflection point and local extremum at that point at the same time.
A: Geometrically speaking, an inflection point is a point where the curve crosses its tangent line. The equation of the tangent line at $x_0$ is $y = f(x_0) + (x - x_0)f'(x_0)$. This is also true if $f'(x_0) = 0$. If the first non-zero derivative after the first one has odd order, then one has an inflection point (assuming such a derivative exists).
