given f(x)=3sec^2(x) , write the equation of the line tangent to y=f(x) when x=11pi/6 PLEASE ANSWER IN Y-Y1=M(X-X1) FORM

Answer 1

The slope of the tangent is the first derivative of the the function,

f (x) =3 sec ² (x)

f ' (x) = 3* 2* sec (x) * sec(x) tan(x)

= 6 sec² (x) tan(x)

at x = 11pi/6

f ' (x) = 6 sec² (11pi/6) tan(11pi/6)

= - 4.6

To write the equation of the line we need to know the slope and a point in the line,

for x₁= 11pi/6 , y₁ = 3 sec ² (x) = 3 sec ² (11pi/6) = 4

y - y₁ = m ( x - x_{1 )}

y - 4 = -4.6 ( x - 11pi/6 )