amy0799
  • amy0799
Let f(x) = sinxcosx + 0 for
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amy0799
  • amy0799
\[0 \le x \le \frac { 3\pi }{ 2} \] find all the values for which f'(x)=1
campbell_st
  • campbell_st
the question really contains + 0....?
amy0799
  • amy0799
oops its suppose to be sinxcosx+x

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More answers

campbell_st
  • campbell_st
ok.... so you need the product rule to do the sin(x)cos(x) or you can use a double angle substitution... do you know sin(2x) = 2 sin(x)cos(x)..?
amy0799
  • amy0799
ill do the product rule f'(x)=cosxcosx+cosx(-sinx)+1
campbell_st
  • campbell_st
well not quite u = sin(x) v = cos(x) u' = cos(x) v' = -sin(x) so the derivative is \[\frac{dy}{dx} = \sin(x) \times - \sin(x) + \cos(x) \times \cos(x) + 1\] or \[\frac{dy}{dx} = \cos^2(x) - \sin^2(x) + 1\] does that make sense
amy0799
  • amy0799
ooh I see what I did wrong
campbell_st
  • campbell_st
so the derivative is equal to 1 then \[1 = \cos^2(x) - \sin^2(x) + 1~~~or~~~0 = \cos^2(x) - \sin^2(x)\] I'd now make a substitution of \[\sin^2(x) = 1 - \cos^2(x)\] so you have \[0 = \cos^2(x) -(1 - \cos^2(x) \] or \[0 = 2\cos^2(x) - 1\] now you should be able to solve for x
amy0799
  • amy0799
what do I put in x? 3pi/2? @campbell_st
campbell_st
  • campbell_st
no you solve so \[1 = 2\cos^2(x)\] then \[\frac{1}{2} = \cos^2(x)\] so \[\pm \frac{1}{\sqrt{2}} = \cos(x)\] this is an exact value.... that is equal to \[x = \frac{\pi}{4}\] in the 1st quadrant so there are angles in the 2nd quadrant \[\pi - \frac{\pi}{4}\] and 3rd quadrant \[\pi + \frac{\pi}{4}\] just simplify the angles in the 2nd and 3rd quadrants by writing them as improper fractions.
amy0799
  • amy0799
how did you get pi/4? @campbell_st
campbell_st
  • campbell_st
because I know \[\frac{1}{\sqrt{2}} = \frac{\pi}{4}\] its called an exact value, you may know it with a rationalized denominator as \[\frac{\sqrt{2}}{2} = \frac{\pi}{4}\]
campbell_st
  • campbell_st
but if you want to check, use a calculator and just enter \[\cos^{-1}(\frac{1}{\sqrt{2}})\]
amy0799
  • amy0799
ooh ok so what do you mena by simplify the angles in the 2nd and 3rd quadrants by writing them as improper fractions?
campbell_st
  • campbell_st
well the angles are normally written using fractions so 3rd quadrant \[\pi + \frac{\pi}{4} = \frac{5\pi}{4}\] you need to work out the 2nd quadrant angle

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