From Wikipedia: a critical point or stationary point of a differentiable function of a single real variable, #f(x)#, 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(x) = x tan x#, we first need to compute #f'(x)# then find all the #x#-values such that #f'(x)=0#. Yet due to it being a trigonometric function, you'll need to be fluent at trigonometry too.
Using the product rule,
#d/(dx) (x tan x) = (d/(dx) x) tan x + x (d/(dx) tan x)#
#f'(x) = tan x + x sec^2 x #
#f'(x) = (sin x)/(cos x) + x/(cos^2 x) #
#f'(x) = (sin x cos x + x)/(cos^2 x) #
When #f'(x)=0#:
#sin x cos x + x = 0#
#1/2 sin 2x + x = 0#
#sin 2x + 2x=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 #(0, 0)#
