Studying the function $f(x) = x^4-6x^2$ using derivatives: minima, maxima, inflection, concavity (I know this is my second question today, but I'm explaining what I'm doing so I hope it's okay)

Consider the graph of $f(x) = x^4-6x^2$.


  a) Find the relative maxima and minima (both x and y coordinates).


I actually forgot how to do this.. not the point of my question though.


  b) Find the coordinates of the point(s) of inflection.


So for this, you take the derivative of the derivative, correct?

$f'(x) = 4x^3-12x$

$f''(x) = 12x^2-12.$

$f''(x)=0 \Leftrightarrow 12x^2 = 12 \Leftrightarrow x = -1, 1.$

So the points are $(-1,5)$ and $(1,5)$ (y coordinate from plugging back into the original)?


  c) Determine the interval(s) on which the function is increasing.


When I set the derivative equal to zero to find the potential turning points, I get 0, rad 3, and negative rad 3. Forgot what to do from here. Not important.


  d) Determine the interval(s) on which the function is concave up.


I get $(-\infty,-1) \cup (1,\infty)$

*Answer only what you want. Any help is appreciated!

Source: http://online.math.uh.edu/apcalculus/exams/AB_SECTION_II_version_1.pdf - #4
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 $f''(x)&gt;0$
after that find $f'(x)=0$ here you get two points.

Then evaluate these two points in $f''(x)$ for each of these two points.

then you get one negative -ve(local maximum),put put this value in f(x).

then other you will get +ve positive (local minimum) value after putting in f''(x). put this value in f(x).

Note: put the x-values in f(x), not that positive and negative value.