Thread: Cos derivative problem

1. Cos derivative problem

Hey i was having a problem with a hw problem
the function f(x)=(cosx)^2 +2cosx over one complete period beginning with x=0
First we have to find all values of x in this period when f(x)=0.
I know the period is 0-2π by looking at the graph however I don't know how to figure it out analyitically though. So i found the zeroes but again i dont know how to do it analytically because i am not got with sin and cos
Second we have to find all values of x in the period at which the function has a minimum.
3rd we have to find when the period is concave up but i don't know how to find the 2nd derivative of a cos

2. $\displaystyle cos^{2}(x)+2cos(x)=0$

The trick to use is let $\displaystyle u=cos(x)$, then we get a quadratic

$\displaystyle u^{2}+2u=0$

$\displaystyle u=0, \;\ -2$

$\displaystyle cos(x)=0, \;\ cos(x)=-2$

-2 is non real, so let's go with 0.

$\displaystyle cos(\frac{\pi}{2})=0$

Therefore, we have f(x)=0 at

$\displaystyle C{\pi}-\frac{\pi}{2}$

Where C is an integer.

To find the second derivative of cos(x):

$\displaystyle f(x)=cos(x), \;\ f'(x)=-sin(x), \;\ f''(x)=-cos(x), \;\ f'''(x)=sin(x)$
$\displaystyle f''''(x)=cos(x)$. See?. It repeats.

3. Originally Posted by calc_help123
Hey i was having a problem with a hw problem
the function f(x)=(cosx)^2 +2cosx over one complete period begining with x=0
First we have to fun all values of x in this period when f(x)=0.
I know the period is 0-2π by looking at the graph however I dont know how to firgure it out analyitically though. So i found the zeroes but again i dont know how to do it analytically bc i am not got with sins and cos
Second we have to find all values of x in the period at which the function has a minimum.
3rd we have to find when the period is concave up but i dont know how to find the 2nd derivative of a cos
$\displaystyle f(x) = \cos^2 x + \cos x$

a) for f(x) = 0,

$\displaystyle 0 = \cos^2 x + \cos x$

$\displaystyle 0 = \cos x(\cos x +2)$

$\displaystyle \Rightarrow \cos x = 0\;\;OR\;\; \cos x +2 =0$

$\displaystyle \Rightarrow \cos x = 0\;\;OR\;\; \cos x \ne -2, \;\; because\;\;-1\le \cos x \le 1$

now finish it

b) For the values of x in which f(x) is minimum, f'(x) = 0

$\displaystyle f'(x) = 2\cos x (-\sin x)+(-\sin x)$

$\displaystyle f'(x) = -\sin x (2\cos x +1)$

$\displaystyle -\sin x (2\cos x +1)=0$

$\displaystyle -\sin x=0\;\;OR\;\;2\cos x +1=0$

$\displaystyle \sin x =0 \;\;OR\;\;\cos x =\frac{-1}{2}$

Now finish it.

c) second derivative, use product rule in $\displaystyle f'(x) = -\sin x (2\cos x +1)$

$\displaystyle f"(x) = -\cos x (2\cos x +1)-\sin x (-2\sin x)$

$\displaystyle f"(x) = -2\cos^2 x -\cos x -2\sin^2 x$

$\displaystyle f"(x) = -2(\cos^2 x + \sin^2 x) - \cos x$

$\displaystyle f"(x) = -2 -\cos x$

Now finish it.

Did you get it now???

4. $\displaystyle f(x) = \cos^2 x + \cos x$
I am sorry Shyam but it is $\displaystyle f(x) = \cos^2 x + 2\cos x$ not $\displaystyle f(x) = \cos^2 x + \cos x$. That changes the first and second derivative drastically doesn't it?

5. so the second derivative is $\displaystyle 2sinx^2-2cosx^2-2cos$?

and the zeros are at 1.0471976, $\displaystyle \pi$, and 5.2359878 ? Can those numbers be made into more exact, easier numbers? maybe in terms of $\displaystyle \pi$? or square roots?

This problem is almost done! Thanks everyone!

6. $\displaystyle f(x) = \cos^2{x} + 2\cos{x}$

$\displaystyle f'(x) = -2\cos{x}\sin{x} - 2\sin{x} = -\sin(2x) - 2\sin{x}$

$\displaystyle f''(x) = -2\cos(2x) - 2\cos{x} = -2[\cos(2x) + \cos{x}] = -2(2\cos^2{x} + \cos{x} - 1)$

the last expression will also factor ...

$\displaystyle f''(x) = -2(2\cos{x} - 1)(\cos{x} + 1)$

$\displaystyle f''(x) = 0$ at $\displaystyle x = \frac{\pi}{3} \, , \, x = \frac{5\pi}{3} \, , \, x = \pi$

7. Thanks skeeter, shyam, galactus and calc_help123!