What is the equation of the line tangent to #f(x)=2 tan(3pix +4)# at #x=1#?

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Answer 1 · 2016-01-16T08:38:51

Tangent Line

#y=6 pi(sec^2 4)x -6 pi* sec^2 4 + 2*tan 4#

The given
#f(x)=2 tan (3 pi x+4)# at #x=1#

#f(1)=2 tan(3 pi *1+4)#
#f(1)=2 tan 4#

The point of Tangency #x_1=1# Abscissa
#y_1= 2 tan 4= 2.31564# Ordinate

The Slope
#f' (x) =2 sec^2(3 pi x+4)* 3 pi#
#f' (x) = 6 pi* sec^2 (3 pi x +4)#
slope #m=6 pi sec^2 (3 pi * 1 +4)#
slope #m=6 pi sec^2 4#

Tangent Line

#y-y_1=m*(x-x_1)#

#y-2 tan 4 = 6 pi sec^2 4 (x - 1)#

#y= 6 pi (sec^2 4)x -6 pi sec^2 4 + 2 tan 4#