# Find the smallest positive x-value where f(x)=x+4sin(2x) has a horizontal tangent line?

## must be in exact value. thanks!

Oct 29, 2017

$x = \frac{1}{2} {\cos}^{-} 1 \left(- \frac{1}{8}\right)$

#### Explanation:

First, we have to understand what it means to have a horizontal tangent line.

We know that the slope of the tangent line to a function is determined by the derivative of the function. If a line is horizontal, its slope is zero.

So, we need to find the smallest value of $x$ where the derivative of the function is $0$.

$f \left(x\right) = x + 4 \sin \left(2 x\right)$

The derivative of $x$ is $1$. To find the derivative of $4 \sin \left(2 x\right)$, we need to consider the derivative of the sine function $\sin \left(2 x\right)$. Well, we know that the derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$. Then, by the chain rule, the derivative of $\sin \left(g \left(x\right)\right)$ is $\cos \left(g \left(x\right)\right) \cdot g ' \left(x\right)$. That is, we take the "cosine" version of the function, then multiply by the derivative of the function on the inside.

In this case, the derivative of $\sin \left(2 x\right)$ will be $\cos \left(2 x\right) \cdot \text{derivative of } 2 x$, which is $2 \cos \left(2 x\right)$. Thus, the derivative of $4 \sin \left(2 x\right)$ is $4 \cdot 2 \cos \left(2 x\right) = 8 \cos \left(2 x\right)$. Then:

$f ' \left(x\right) = 1 + 8 \cos \left(2 x\right)$

We need to find when the derivative equals $0$, which is when:

$0 = 1 + 8 \cos \left(2 x\right)$

Rearranging,

$\cos \left(2 x\right) = - \frac{1}{8}$

That is,

$2 x = {\cos}^{-} 1 \left(- \frac{1}{8}\right)$

The range of the ${\cos}^{-} 1 \left(x\right)$ function is $\left[0 , \pi\right]$, so we know that ${\cos}^{-} 1 \left(- \frac{1}{8}\right)$ will give us the smallest positive value that applies in this situation.

Thus:

$x = \frac{1}{2} {\cos}^{-} 1 \left(- \frac{1}{8}\right)$

You said you wanted an exact value, so I'll leave it like this.