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

must be in exact value.
thanks!

1 Answer
Oct 29, 2017

x=1/2cos^-1(-1/8)x=12cos1(18)

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 xx where the derivative of the function is 00.

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

The derivative of xx is 11. To find the derivative of 4sin(2x)4sin(2x), we need to consider the derivative of the sine function sin(2x)sin(2x). Well, we know that the derivative of sin(x)sin(x) is cos(x)cos(x). Then, by the chain rule, the derivative of sin(g(x))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.