Find points where f(x)= 4sinX+2x has a horizontal tangent 10<=x<=2(pi)
B brian ravenell New member Joined Sep 24, 2006 Messages 1 Sep 24, 2006 #1 Find points where f(x)= 4sinX+2x has a horizontal tangent 10<=x<=2(pi)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Sep 24, 2006 #2 Re: help me Hello, Brian! Find points where f(x) = 4sinx + 2x\displaystyle f(x)\:=\:4\sin x\,+\,2xf(x)=4sinx+2x has a horizontal tangent, 0 ≤ x ≤ 2π\displaystyle 0\,\leq\,x\,\leq\, 2\pi0≤x≤2π Click to expand... Horizontal tangents occur where f′(x) = 0\displaystyle f'(x)\,=\,0f′(x)=0 We have: 4cosx + 2 = 0 ⇒ cosx = −12\displaystyle \:4\cos x\,+\,2\:=\:0\;\;\Rightarrow\;\;\cos x\,=\,-\frac{1}{2}4cosx+2=0⇒cosx=−21 Therefore: x = 2π3, 4π3\displaystyle \:x\:=\:\frac{2\pi}{3},\:\frac{4\pi}{3}x=32π,34π
Re: help me Hello, Brian! Find points where f(x) = 4sinx + 2x\displaystyle f(x)\:=\:4\sin x\,+\,2xf(x)=4sinx+2x has a horizontal tangent, 0 ≤ x ≤ 2π\displaystyle 0\,\leq\,x\,\leq\, 2\pi0≤x≤2π Click to expand... Horizontal tangents occur where f′(x) = 0\displaystyle f'(x)\,=\,0f′(x)=0 We have: 4cosx + 2 = 0 ⇒ cosx = −12\displaystyle \:4\cos x\,+\,2\:=\:0\;\;\Rightarrow\;\;\cos x\,=\,-\frac{1}{2}4cosx+2=0⇒cosx=−21 Therefore: x = 2π3, 4π3\displaystyle \:x\:=\:\frac{2\pi}{3},\:\frac{4\pi}{3}x=32π,34π