# How do you find the points where the graph of the function y=tan(x)-x has horizontal tangents?

Oct 26, 2016

x=kpi; k in ZZ

#### Explanation:

${f}^{'} \left(x\right) = \frac{1}{\cos} ^ 2 x - 1$
${f}^{'} \left(x\right) = 0$
when
${\cos}^{2} x = 1$
or
$\cos x = \pm 1$