# Thread: Derivative to Find Horizontal Tangent

1. ## Derivative to Find Horizontal Tangent

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?

2. 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

3. 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?

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

5. 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.

6. 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 $x= \frac{11\pi}{6}$

7. 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?