# Thread: Finding values of x that have have horizontal tangent.

1. ## Finding values of x that have have horizontal tangent.

Hi guys! I've been working on this for a while:

I've restarted twice and have gotten to the same step with both:

-6 sinx - sin^2x - 5 = 0

I know I need to set the y(prime) equal to 0, so I used the quotient rule and then the -sin^2 x - cos^2 x = -1 identity and seem to have hit some sort of cul-de-sac, as it were! Any guidance is appreciated!

2. $y = \frac{\cos{x}}{2+\sin{x}}$

$y' = \frac{(-\sin{x})(2+\sin{x}) -(\cos{x})(\cos{x})}{(2+\sin{x})^2}$

$y' =\frac{-2\sin{x} -\sin^2{x} - \cos^2{x}}{(2+\sin{x})^2}$

$y' =\frac{-2\sin{x} -\sin^2{x} -\cos^2{x}}{(2+\sin{x})^2}$

$y' =\frac{-2\sin{x}-1}{(2+\sin{x})^2)}$

Now set the numerator equal to 0 and solve to find your horizontal tangents

3. Originally Posted by 11rdc11
$y = \frac{\cos{x}}{2+\sin{x}}$

$y' = \frac{(-\sin{x})(2+\sin{x}) -(\cos{x})(\cos{x})}{(2+\sin{x})^2}$

$y' =\frac{-2\sin{x} -\sin^2{x} - \cos^2{x}}{(2+\sin{x})^2}$

$y' =\frac{-2\sin{x} -\sin^2{x} -\cos^2{x}}{(2+\sin{x})^2}$

$y' =\frac{-2\sin{x}-1}{(2+\sin{x})^2)}$

Now set the numerator equal to 0 and solve to find your horizontal tangents
Hey rdc, I had gotten to that last step that you showed but I didn't know you could just set the numerator to 0 (makes this problem a lot easier). What happens to the denominator in this situation?

4. Zero divided by anything else will always equal zero, which in your case will cause the gradient to equal zero.

So if you set your numerator to zero, the denominator will not matter.

5. Ok, so you multiply both sides by the denominator and that's why the denominator basically disappears? I've never heard of the word gradient in this context. This all might sound trivial haha but I'm interested in learning why.

6. Gradient is another way of saying slope. When you take multivariable calculus you will also learn a new defintion for gradient. And nope the denominator does matter. If you set that equal to 0 you can find vertical tangents.

$\lim_{x \to a} \bigg|f'(x)\bigg| = \infty$