Trigonomic Function

[fgmango]

The function is f (x) =3sinx+cos2x

f'(x)=3cosx-2sin2x

f"(x)=-3sinx-4cos2x

Is this right?

 

[stapel]

Looks good to me.

Eliz.

 

[fgmango]

I'm having trouble finding the x-int

f"(x)=-3sinx-fcos2x = -sinx(3+4sinx)

-sinx=0 3+4sinx=o
sinx=0 4sinx=-3
sinx=-3/4

Is this right?

 

[stapel]

What is the "f" in "-3sin(x) - fcos(2x)"?

How are you getting that 4cos(2x) equals 4sin²(x)?

Eliz.

 

[fgmango]

That f is suppose to be a 4.

I was trying to substite in cos2x=1+sinx

 

[stapel]

You might want to check your identities: cos(2x) = 1 + sin²(x).

Eliz.

 

[fgmango]

Right my mistake. Thats how I was trying to factor out -sinx.

 

[soroban]

Hello, fgmango!

I'm having trouble finding the x-int

No wonder! . . . You're solving: $\displaystyle f''(x)\,=\,0$
. . You're looking for inflection points!

$\displaystyle f''(x)\:=\:-3\cdot\sin(x)\,-\,4\cdot\cos(2x)\:=\:0$

Then: .-$\displaystyle 3\cdot\sin(x)\,-\,4[1\,-\,2\cdot\sin^2(x)]\:=\:0$

We have a quadratic: .$\displaystyle 8\cdot\sin^2(x)\,-\,3\cdot\sin(x)\,-\,4\:=\:0$

. . with roots: .$\displaystyle \sin(x)\:=\:\frac{3\,\pm\,\sqrt{137}}{16}$


Inflection points occur at: .$\displaystyle x\;=\;\arcsin\left(\frac{3\,\pm\,\sqrt{137}}{16}\right)$

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If you're really looking for $\displaystyle x$-intercepts, set $\displaystyle f(x)\,=\,0.$

We have: .$\displaystyle 3\cdot\sin(x)\,+\,\cos(2x)\:=\:0$

Then: .$\displaystyle 3\cdot\sin(x)\,+\,[1\,-\,2\cdot\sin^2(x)]\:=\:0$

We have a quadratic: .$\displaystyle 2\cdot\sin^2(x)\,-\,3\cdot\sin(x)\,-\,1\:=\:0$

. . with roots: .$\displaystyle \sin(x)\;=\;\frac{3\,\pm\,\sqrt{17}}{4}$


$\displaystyle x$-intercepts occur at: .$\displaystyle x\:=\:\arcsin\left(\frac{3\,\pm\,\sqrt{17}}{4}\right)$
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