For what values of x does f(x) = 4x + 8 sin x have a horizontal tangent line?

Aug 4, 2018

$x = 2 k \pi \pm 2 \frac{\pi}{3} , k \in \mathbb{Z}$.

Explanation:

Recall that the slope of tangent line (tgt.) is given by $f ' \left(x\right)$.

So, if the tgt. is horizontal, its slope, i.e., $f ' \left(x\right) = 0.$

Now, $f \left(x\right) = 4 x + 8 \sin x \Rightarrow f ' \left(x\right) = 4 + 8 \cos x$.

$\therefore f ' \left(x\right) = 0 \Rightarrow 4 + 8 \cos x = 0 \Rightarrow \cos x = - \frac{4}{8} = - \frac{1}{2}$.

But, $\cos \theta = \cos \alpha \Rightarrow \theta = 2 k \pi \pm \alpha , k \in \mathbb{Z}$.

Therefore, $\cos x = - \frac{1}{2} = \cos \left(\pi - \frac{\pi}{3}\right) = \cos \left(2 \frac{\pi}{3}\right) ,$

$\Rightarrow x = 2 k \pi \pm 2 \frac{\pi}{3} , k \in \mathbb{Z}$.