given f(x)=2sec^2(x) , write the equation of the line tangent to y=f(x) when x=7pi/4 PLEASE PUT ANSWER IN y-y1=m(x-x1) FORM ⚠️⚠️⚠️PLEASE PUT ANSWER IN y-y1=m(x-x1) FORM⚠️⚠️⚠️

Question

PLEASE PUT ANSWER IN Y-Y1=M(X-X1) FORM!!!!!

Answer 1

f(7π/4) = 2sec²(7π/4) = 2(√2)² = 4; tangent point is (7π/4, 4).

f'(x) = 2secxsecxtanx = 2sec²xtanx; Slope of tangent line is f'(7π/4) = 2sec²(7π/4)tan7π/4 = - 4.

So, equation of tangent line is y - 4 = - 4(x - 7π/) or y - 4 = - 4x + 7π

Answer 2

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

f (x) =2 sec ² (x)

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

= 4 sec² (x) tan(x)

at x = 7π/4

f ' (x) = 4 sec² ( 7π/4) tan( 7π/4)

= 4 * 2 * (-1) = -8

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

for x₁= 7π/4 , y₁ = 2 sec ² (x) =2 sec ² ( 7π/4) = 2 * 2 = 4

y - y₁ = m ( x - x_{1 )}

y - 4 = -8 ( x - 7π/4 )