# Question f73ad

Aug 10, 2015

$\left(0 , 0\right)$

#### Explanation:

From Wikipedia: a critical point or stationary point of a differentiable function of a single real variable, $f \left(x\right)$, is a value ${x}_{0}$ in the domain of $f$ where its derivative is 0: f′(x_0) = 0#

Thus, to find the critical points of $f \left(x\right) = x \tan x$, we first need to compute $f ' \left(x\right)$ then find all the $x$-values such that $f ' \left(x\right) = 0$. Yet due to it being a trigonometric function, you'll need to be fluent at trigonometry too.

Using the product rule,
$\frac{d}{\mathrm{dx}} \left(x \tan x\right) = \left(\frac{d}{\mathrm{dx}} x\right) \tan x + x \left(\frac{d}{\mathrm{dx}} \tan x\right)$
$f ' \left(x\right) = \tan x + x {\sec}^{2} x$
$f ' \left(x\right) = \frac{\sin x}{\cos x} + \frac{x}{{\cos}^{2} x}$
$f ' \left(x\right) = \frac{\sin x \cos x + x}{{\cos}^{2} x}$

When $f ' \left(x\right) = 0$:
$\sin x \cos x + x = 0$
$\frac{1}{2} \sin 2 x + x = 0$
$\sin 2 x + 2 x = 0$

The solution of this equation, if you plot the graph, is $x = 0$

graph{sin(2x)+2x [-20, 20, -10, 10]}

Thus, only $x = 0$ gives us a critical point. The critical value, then, is $y = x \tan x = 0 \tan 0 = 0$

The coordinates of the critical point is $\left(0 , 0\right)$