If $f(x)$ has a local max or min at $x=0$, then $f'(0)=0$. Now suppose $f(x)$ has an inflection point at $x=0$.
(case 1) $f'(x)$ is increasing on $(-\epsilon,0]$ and decreasing on $[0,\epsilon)$. Then that means $f'(x) \leq 0$ on $(-\epsilon,\epsilon)$. So $f(x)$ is strictly decreasing on $(-\epsilon,\epsilon)$ so there can't possibly be a local max/min at $x=0$.
(case 2) $f'(x)$ is decreasing on $(-\epsilon,0]$ and increasing on $[0,\epsilon)$. Then that means $f'(x) \geq 0$ on $(-\epsilon,\epsilon)$. So $f(x)$ is strictly increasing on $(-\epsilon,\epsilon)$ so there can't possibly be a local max/min at $x=0$.
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Lets consider $$f(x, y) = x^3 - y^3 - 12xy -63x + 63y$$
Taking both partial derivate of $x$ and $y$ and making them equal $0$,
$$\frac{\partial}{\partial x} f(x, y) = 3x^2-12y-63 = 0, \quad (1)$$ and $$\frac{\partial}{\partial y} f(x, y) = -3y^2 -12x +63x = 0, \quad (2)$$
Clearing $y$ from equation $(1)$ and substituting in equation $(2)$, $$y=\frac{3x^2-63}{12} \Longrightarrow -3\Big(\frac{3x^2-63}{12}\Big)^2-12x+63=-\frac{3x^4}{16}+\frac{63x^2}{8}-12x+\Big(63-\frac{1323}{16}\Big)=0$$ which is a fourth degree polynomial, then $$x\in\{-7, -1, 3, 5\}$$ Using this set of points for $x$ on equation $(1)$ (or $(2)$, it's the same), were have the following set of stationary points, $$A = \{(-7, 7), (-1, -5), (3,-3), (5, 1)\}$$
Let's consider the following Theorem.
Let $$D=\frac{\partial^2}{\partial x^2} f(a, b) \cdot \frac{\partial^2}{\partial y^2} f(a, b) -\Big(\frac{\partial^2}{\partial x \partial y} f(a, b)\Big)^2, \quad (a, b) \in A$$ then,
$1)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) > 0$, then $f$ has a minimum at $(a, b)$.
$2)$ If $D>0$ and $\frac{\partial^2}{\partial x^2} f(a, b) < 0$, then $f$ has a maximum at $(a, b)$.
$3)$ If $D<0$, then $f$ has a saddle point at $(a, b)$.
$4)$ If $D=0$, then we have no conclusion for $f$ at $(a, b)$.
Using this Theorem with the points of the set $A$, we conclude that, $(-7, 7)$ is maximum of $f$, $(3, -3)$ is minimum of $f$ and both $(-1, -5)$ and $(5, 1)$ are saddle points of $f$.
Best Answer
a) Min/Max: calculate the derivative of the function and equal it to zero to find the x-coordinate of the extreme points of the function. Then calculate the second derivative and evaluate it on these extreme values of $x$. If the 2nd derivative is positive, then the slope of the tangent is growing and thus we have a minimum at $(x, f(x))$. if the 2nd derivative is negative; we have a maximum $(x, f(x))$
b) That's right.
c) function opens up when $f''(x)>0$