anonymous
  • anonymous
24: find Point of Inflection and discuss concavity of graph of func. 40: find all relative extrema, use 2nd derivative test where applicable. 58: Water is running into the vase shown @ a constant rate. (a) graph depth of water in vase as a function of time. (b) Does the function have any Extrema? Explain. (c) Interpret the PI (Point of Inflection) of the graph of d.
Calculus1
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
My work:
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anonymous
  • anonymous
@Shadowys can you help please? :O
anonymous
  • anonymous
Or anyone really :P

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anonymous
  • anonymous
lol, um, meanwhile, can you type the original eq.?
anonymous
  • anonymous
okay
anonymous
  • anonymous
#24: f(x) = sinx+cosx [0, 2Pi] #40: f(x) 2sinx + cos2x [0, 2Pi] I think mostly i'm having issues with just finding the derivative of trig functions. Oh, and my friend helped me with number 58 so it's all cool.
anonymous
  • anonymous
for question 40, for the test, you let f'(x) = 0 and solve for x. then, you apply the x-values into f''(x) to test for maxima or minima
anonymous
  • anonymous
point of inflexion in 24 is found by letting f''(x)=0. also, you'll need the double angle formulas.
anonymous
  • anonymous
yeah, i know how to = 0 but I don't know how to solve the trig to find the x :(
anonymous
  • anonymous
|dw:1354600377935:dw|
anonymous
  • anonymous
lol okay, i'll do one as an example. \(-sin x- cos x=0\) \(cos x+ sin x=0\) applying the \(Rcos(x-\alpha)\) form, where \(R=\sqrt{1+1}\), \(\alpha = tan^{-1} 1=\pi/4\) \(\sqrt2 cos (x-\pi/4)=0\) solve for x.
anonymous
  • anonymous
i don't think i've ever seen the Rcos(x−α) form before. :O
anonymous
  • anonymous
hmm, another way is letting sin x = cos(pi\2 -x) but this is much more complicated, but yes, this is the way to solve these type of eqs. for angles with the same cosine, i.e. \(\cos \theta= \cos \alpha\),\( \theta = 2n\pi \pm \alpha\)
anonymous
  • anonymous
okay, thanks for your help!
anonymous
  • anonymous
you're welcome :)

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