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

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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 $(pi/4,0)$ ,

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

$y-0=0$

$y=0$

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