anonymous
  • anonymous
for the curve with the equation y = 3x^2 -x^3, find the values of x for which y is decreasing
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Learn the standard form for a quadratic equation. The equation is y = ax^2 + bx + c. The letters a, b and c represent coefficients. Determine whether the value of the a coefficient of your equation is equal to zero. For example, your equation is y = 7x + 11. The "a" coefficient is equal to 0: y = 0x^2 + 7x + 11. Your equation represents a line. determine whether the value of the "a" coefficient of your equation is not equal to zero. For example, your equation is y = 3x^2 + 7x + 11. The "a" coefficient is not equal to 0; it is equal to 3.
anonymous
  • anonymous
ummm my question is y = 3x^2 -x^3 which can be written as y =-x^3 + 3x^2
anonymous
  • anonymous
i don't understand why what you just wrote has anything to do with solving the question , other than telling me about basic quadratic functions

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anonymous
  • anonymous
y'=6x-3x^2 -3x^2+6x=0 -3x(x-2)=0 x=0, 2 divide the domain into 3 intervals (-infinity, 0), (0, 2), and (2, infinity) test a sample from each domain and find the sign of the test. if the sign is negative, its decreasing
anonymous
  • anonymous
from each interval*
anonymous
  • anonymous
ya in simpler lang. @dock... is correct
anonymous
  • anonymous
no rohangrr, you were completely off, but thanks for trying. i'll give you are medal anyways
anonymous
  • anonymous
DOCKWORKER can i just use 2nd derivative to test instead of split it into 3 intervals?
anonymous
  • anonymous
sure, but its quite easy using 1st derivative test in this case
anonymous
  • anonymous
oh i see, so if i did use second derivative when i test it for x=0, second derivative is 6 which is > 0 but for x=2, second derivative is -6 which is <0 hence y is decreasing at x =2 is this how i would do it dockland?
anonymous
  • anonymous
oh sorry i mean dockworker not dockland. my apologies!
anonymous
  • anonymous
2nd derivative test would tell you if there was a max or min at that point, not if it was increasing or decreasing
anonymous
  • anonymous
oh i see, so could you show me how i would do it using second derivative please?
anonymous
  • anonymous
y''=-6x+6 -6(0)+6=6, so f(0) is a local minimum -6(2)+6=-6, so f(6) is a local max for x=0 to produce a min, the graph would switch from decreasing before x=0 to increasing after x=0. for x=2 to produce a max, the graph would be increasing before x=2 to increasing after x=2
anonymous
  • anonymous
so its decreasing on the interval [-infinity, 0)U(2, infinity)
anonymous
  • anonymous
(-infinity, 0)U(2, infinity)**
anonymous
  • anonymous
oh i see! THANK YOU DOCKWORKER! your help is greatly appreciated!
anonymous
  • anonymous
you're welcome

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