What is the equation of the line tangent to #f(x)=x ^2cos^2(2x) # at #x=pi/4#?

Question

980 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-17T10:42:54

#y=0#

#f(x)= x^2(cos 2x)^2#

graph{x^2(cos(2x))^2 [-0.0585, 1.44, -0.295, 0.4544]}

#f'(x)=x^2times2(cos2x)times-2(sin2x)+(cos2x)^2times2x#
#f'(x)=-4x^2sin2xcos2x+2x(cos2x)^2#

At #x=pi/4#,

#f'(pi/4)=-4(pi/4)^2sin(2timespi/4)cos(2timespi/4)+2(pi/4)(cos2timespi/4)^2#
#f'(pi/4)=0#

Equation of line at #(pi/4,0)#,

#(y-0)=0(x-pi/4)#

#y-0=0#

#y=0#