chris215
  • chris215
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Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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DanJS
  • DanJS
solve for the derivative f '(x) = 0 there it may change inc/dec or dec/inc
DanJS
  • DanJS
test each one to see if the sign changes from one side of tht point to the other
chris215
  • chris215
ok what I first did was find the first and second derivatives -4x^3+4x 2nd -12x^2+4

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mathmale
  • mathmale
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.
mathmale
  • mathmale
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?
chris215
  • chris215
−4x^3+4x=0 x= 0,1,-1
chris215
  • chris215
when I evaluate the second derivative at each of those critical values I got 0 tho
mathmale
  • mathmale
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.
chris215
  • chris215
I dont get it
anonymous
  • anonymous
google (-x^4 + 2x^2)
DanJS
  • DanJS
inc / dec , is all they want, maximum or minimum will change from one to the other
DanJS
  • DanJS
local max/min
welshfella
  • welshfella
can you find the derivative of -x^4 = 2x^2 ?
DanJS
  • DanJS
just test an x on eithe rside of the critical value, see if it changes in the first derivative
DanJS
  • DanJS
f'(x) will change sign - to + at a min or + to - at a max
welshfella
  • welshfella
oh sorry you have already
SolomonZelman
  • SolomonZelman
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.
DanJS
  • DanJS
the second derivative about the concavirty, shows inflections
DanJS
  • DanJS
|dw:1449358986708:dw| test a value in each interval , defined by those crit values
DanJS
  • DanJS
|dw:1449359236741:dw|
DanJS
  • DanJS
increasing - decreasing - increasing - decreasing name the intervals they want, where it changes from decreasing tp increasing
DanJS
  • DanJS
general shape, for those intervals gives |dw:1449359631182:dw|
chris215
  • chris215
ok I got 0 and -1
DanJS
  • DanJS
just the 0. x = 1, the function changes increasing to dec,

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