Given $y = 2x + \sqrt{3}\cos{x}$, find all horizontal tangents and write a general solution for x.
To find horizontal tangents, we derive and equal to 0:
$y' = 2 – \sqrt{3} \sin{x}$ ; $y'=0$
$2 – \sqrt{3} \sin{x} = 0$
$\sin{x}= \frac{2}{\sqrt{3}}$
$x = \arcsin(\frac{2}{\sqrt{3}})$
However, $\arcsin(\frac{2}{\sqrt{3}})$ is not a real number.
Upon graphing, I found the horizontal tangents to be $x = \frac{π}{2} + πn$ where n is an integer.
But I don't know how to find the x values without graphing.
Any ideas? Thank you!
Best Answer
What makes you think the function has real horizontal tangent lines? Unless you meant $f(x)=x \sqrt{3}+ 2 \cos(x)$ it does have real horizontal tangents.
Proceeding with the original function by taking the first derivative: $$\frac{d}{dx}\ (2x+\sqrt{3}\ \cos(x)) = 2 - \sqrt{3}\ \sin(x)$$ Setting it equal to zero:
$$2 - \sqrt{3} \ \sin(x) = 0$$ $$\sin(x) = \frac{2}{\sqrt{3}} = \frac{2}{3} \sqrt{3} \ \approx 1.1547$$
Since the range of any sine function is $-1 \le \sin(x) \le 1$, $1.15$ exceeds $1$, so in conclusion, the function $f(x) = 2x - \sqrt{3} \ \cos(x)$ does not have any real horizontal tangent lines.