Turning points of y = a sin(2x) + 2 cos(x)

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Apr 2010
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1)
Given function: \(\displaystyle y=a\sin(2x)+2\cos(x)\) between \(\displaystyle 0\leq x \leq \pi\)
The function has a stationary point on \(\displaystyle x=\frac{5}{6}\pi\)

A) Calculate a.
B) Find the vertex of the function in this domain.
 

mr fantastic

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1)
Given function: \(\displaystyle y=a\sin(2x)+2\cos(x)\) between \(\displaystyle 0\leq x \leq \pi\)
The function has a stationary point on \(\displaystyle x=\frac{5}{6}\pi\)

A) Calculate a.
B) Find the vertex of the function in this domain.
Start by finding dy/dx. Can you do that?
 

e^(i*pi)

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1)
Given function: \(\displaystyle y=a\sin(2x)+2\cos(x)\) between \(\displaystyle 0\leq x \leq \pi\)
The function has a stationary point on \(\displaystyle x=\frac{5}{6}\pi\)

A) Calculate a.
B) Find the vertex of the function in this domain.
\(\displaystyle y' \left(\frac{5\pi}{6}\right) = 0\)

Use the chain rule to differentiate y.
 
Apr 2010
16
1
i already found the a, a=1
my problem is with B.
i got:
\(\displaystyle y'=2\cos(2x)-2\sin(x)=0\)
but i dont getting \(\displaystyle \frac{5}{6}\pi\) as answer. (Crying)
 

mr fantastic

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Dec 2007
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i already found the a, a=1
my problem is with B.
i got:
\(\displaystyle y'=2\cos(2x)-2\sin(x)=0\)
but i dont getting \(\displaystyle \frac{5}{6}\pi\) as answer. (Crying)
It saves time and effort if you say in your question what you can already do.

For B) your job is to find the value of y when x = pi/6. Where are you stuck?
 
Apr 2010
16
1
in finding the x from: \(\displaystyle y'=2\cos(2x)-2\sin(x)=0\)
 

mr fantastic

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