Question

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

must be in exact value.
thanks!

Answer 1

`x=1/2cos^-1(-1/8)`

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(x)=x+4sin(2x)`

The derivative of `x` is `1`. To find the derivative of `4sin(2x)`, we need to consider the derivative of the sine function `sin(2x)`. Well, we know that the derivative of `sin(x)` is `cos(x)`. Then, by the chain rule, the derivative of `sin(g(x))` is `cos(g(x))*g'(x)`. 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(2x)` will be `cos(2x) * "derivative of " 2x`, which is `2cos(2x)`. Thus, the derivative of `4sin(2x)` is `4*2cos(2x)=8cos(2x)`. Then:

`f'(x)=1+8cos(2x)`

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

`0=1+8cos(2x)`

Rearranging,

`cos(2x)=-1/8`

That is,

`2x=cos^-1(-1/8)`

The range of the `cos^-1(x)` function is `[0,pi]`, so we know that `cos^-1(-1/8)` will give us the smallest positive value that applies in this situation.

Thus:

`x=1/2cos^-1(-1/8)`

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