Intervals, local values, concave

[scrum]

4x^3-48x+4

Input the interval(s) on which f is increasing.

Input the interval(s) on which f is decreasing.

Find the point(s) at which f achieves a local maximum.

Find the point(s) at which f achieves a local minimum.

Find the intervals on which f is concave up.

Find the intervals on which f is concave down.

Find all inflection points.

I'm having problems with this type of problem. It's my own fault really, because i have a program on my computer called grapher, which can graph equations and basically do the problem for me. I've been using it to do problems but then I realized I have no idea what to without just graphing it on a calculator and looking at it. presumably there's a way to do it with the derivatives or second derivative. All i really need is the formula and i'll be able to get the rest. Thanks

 

[galactus]

If \(\displaystyle f''(x_{0})>0\), then f has a relative minimum at \(\displaystyle x_{0}\)

If \(\displaystyle f''(x_{0})<0\), then f has a relative maximum at \(\displaystyle x_{0}\)

If \(\displaystyle f''(x)>0\) on an open interval, then f is concave up on that interval.

If \(\displaystyle f''(x)<0\) on an open interval, then f is concave down on that interval.

If \(\displaystyle f'(x)>0\) for every value of x in an interval, then f is increasing

If \(\displaystyle f'(x)<0\), then decreasing.

If \(\displaystyle f'(x)=0\), then f is constant.

To find the max and mins, set f'(x)=0 and solve for x.

For inflection points, set f''(x)=0 and solve for x. That's where it changes concavity.

For instance, f''(x)=24x. What makes this 0?. Not too many solutions.

\(\displaystyle f'(x)=12x^{2}-48\)

What makes this 0?. Those are your max and min.