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solve for the derivative f '(x) = 0 there it may change inc/dec or dec/inc
test each one to see if the sign changes from one side of tht point to the other
ok what I first did was find the first and second derivatives -4x^3+4x 2nd -12x^2+4
Good. Label these derivatives. Find the three values of x that make the first derivative = to 0. Check each in the 2nd derivative. If the result is concave upward, that's where the graph changes from decreasing to increasing.
You find the x-coordinates of the critical points by finding the 1st derivative. You use the 2nd derivative to determine whether your curve is concave up (which signifies a minimum at that x value) or down (which signifies a max). Questions?
−4x^3+4x=0 x= 0,1,-1
when I evaluate the second derivative at each of those critical values I got 0 tho
If your second derivative is 0, that means the 2nd derivative test won't work, and that you must rely on some other method to determine how your graph is behaving.
I dont get it
google (-x^4 + 2x^2)
inc / dec , is all they want, maximum or minimum will change from one to the other
can you find the derivative of -x^4 = 2x^2 ?
just test an x on eithe rside of the critical value, see if it changes in the first derivative
f'(x) will change sign - to + at a min or + to - at a max
oh sorry you have already
Other Critical Points: If f'(x) is undefined, and f(x) is defined. (Saying some point where f is not differentiable) Interval boundaries are given, then closed (the included) boundaries are also critical points. But, respectively, it is a polynomial, and no interval is given, so these two cases don't apply, but good to know for this kind of problems in general.
the second derivative about the concavirty, shows inflections
|dw:1449358986708:dw| test a value in each interval , defined by those crit values
increasing - decreasing - increasing - decreasing name the intervals they want, where it changes from decreasing tp increasing
general shape, for those intervals gives |dw:1449359631182:dw|
ok I got 0 and -1
just the 0. x = 1, the function changes increasing to dec,