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 2 years ago
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.
 2 years ago
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.

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LotusWings
 2 years ago
Best ResponseYou've already chosen the best response.0@Shadowys can you help please? :O

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1lol, um, meanwhile, can you type the original eq.?

LotusWings
 2 years ago
Best ResponseYou've already chosen the best response.0#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.

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1for question 40, for the test, you let f'(x) = 0 and solve for x. then, you apply the xvalues into f''(x) to test for maxima or minima

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1point of inflexion in 24 is found by letting f''(x)=0. also, you'll need the double angle formulas.

LotusWings
 2 years ago
Best ResponseYou've already chosen the best response.0yeah, i know how to = 0 but I don't know how to solve the trig to find the x :(

LotusWings
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1354600377935:dw

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1lol 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.

LotusWings
 2 years ago
Best ResponseYou've already chosen the best response.0i don't think i've ever seen the Rcos(x−α) form before. :O

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1hmm, 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. \(\cos \theta= \cos \alpha\),\( \theta = 2n\pi \pm \alpha\)

LotusWings
 2 years ago
Best ResponseYou've already chosen the best response.0okay, thanks for your help!
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