Derivative to Find Horizontal Tangent

**NotEinstein** · September 17th 2008, 12:41 PM

f(x)= 2secx + tanx for all x in the interval 0 and 2pi

Am i supposed to take the derivative of f(x), set it equal to 0, and find all points where x satisfies the equation?

**Moo** · September 17th 2008, 12:49 PM

Hi

Originally Posted by NotEinstein

f(x)= 2secx + tanx for all x in the interval 0 and 2pi

Am i supposed to take the derivative of f(x), set it equal to 0, and find all points where x satisfies the equation?

Oh yeah

**NotEinstein** · September 17th 2008, 01:33 PM

I'm breaking this problem down, and am stuck at

$2(sinx/cos^2(x))+1+tan^2(x)$

did I mess up the process somewhere, or am i just needing to go further?

**11rdc11** · September 17th 2008, 01:42 PM

$f'(x) =2 \sec{x} \tan{x} + \sec^2{x}$

**NotEinstein** · September 17th 2008, 01:46 PM

Originally Posted by 11rdc11

$f'(x) =2 \sec{x} \tan{x} + \sec^2{x}$

i got that, but i need to plug in to find points where there is a horizontal tangent.. so now I am trying to simplify this derivative.

**11rdc11** · September 17th 2008, 02:45 PM

Factor out a $\sec{x}$

$\sec{x}(2\tan{x} + \sec{x})= 0$

so now you have

$\sec{x} = 0$ and $2\tan{x} + \sec{x} =0$ and you don't have to worry about sec(x) = 0 since that is not possible

$\frac{2\sin{x}}{\cos{x}}+ \frac{1}{\cos{x}} = 0$

$\frac{2\sin{x} + 1}{\cos{x}}$

$2\sin{x} + 1 = 0$

$2\sin{x} = -1$

$\sin{x} = -\frac{1}{2}$

$x= \frac{7\pi}{6}$ and $x= \frac{11\pi}{6}$

**NotEinstein** · September 17th 2008, 03:37 PM

Originally Posted by 11rdc11

Factor out a $\sec{x}$

$\sec{x}(2\tan{x} + \sec{x})= 0$

so now you have

$\sec{x} = 0$ and $2\tan{x} + \sec{x} =0$ and you don't have to worry about sec(x) = 0 since that is not possible

$\frac{2\sin{x}}{\cos{x}}+ \frac{1}{\cos{x}} = 0$

$\frac{2\sin{x} + 1}{\cos{x}}$

$2\sin{x} + 1 = 0$

$2\sin{x} = -1$

$\sin{x} = -\frac{1}{2}$

$x= \frac{7\pi}{6}$

i followed it all the way til the end. how did you figure that x = 7pi/6?

**11rdc11** · September 17th 2008, 03:39 PM

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