anonymous
  • anonymous
how many points of inflection does the graph y=cosx^2 have on the interval (0, pi)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
how do you know?
experimentX
  • experimentX
double differentiate => y = (cos(x))^2 y' = - 2cos(x)sin(x) = - sin(2x) y'' = - 2 cos(2x) for point of inflection, y''=0 or cos(2x) = 0 x=pi/4, 3pi/4
campbell_st
  • campbell_st
its 2 f(x) = cos^2x f'(x) = -2sin(x)cos(x) f"(x) = -2cos(2x) solving -2cos(2x) = 0 \[2x = \cos^{-1}(0) \] \[2x = \pi\] \[x = \pi/2\] 1st and 2nd quadrant x = \[x =\pi/4, 3\pi/4\]

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anonymous
  • anonymous
isn't it just one because it's on the interval of 0 to pi and 3pi/4 lies in negative x
campbell_st
  • campbell_st
3pi/4 < pi.. so it needs to be included

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