given f(x)=2sec^2(x) , write the equation of the line tangent to y=f(x) when x=7pi/4

Question

has to be in standard form y-y1=m(x-x1)

Answer 1

f(x) =y= f(7pi/4) = 2sec^2x = 2sqr2 when x = 7pi/4

f'(x) = (2sec^2x)' = 2(2)secxsecxtanx = 4sec^2xtanx = f'(7pi/4) = 4(2sqr2)(1) = 8sqr2 = slope of the tangent line when x=7pi/4

tangent line is

y=mx +b, m=8sqr2, solve for b by plugging in the points (7pi/r, 2sqr2)

2sqr2 = 8sqr2 + b

b = 2sqr2 - 8sqr2 =-6sqr2

y= 8sqr2(x) -6sqr2 or about y= 8(1.414)x - 6(1.414) = 11.3x - 8.6

y=11.3x -8.6 is approximately the equation of the tangent line to f(x) when x = 7pi/4