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

Status
Not open for further replies.

#### mathwhat

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

MHF Hall of Fame
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)

MHF Hall of Honor
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.

#### mathwhat

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

MHF Hall of Fame
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?

#### mathwhat

in finding the x from: $$\displaystyle y'=2\cos(2x)-2\sin(x)=0$$

Status
Not open for further replies.