Question

Use linear approximation to estimate sec(44)

Answer 1

Take a look at this: http://is.gd/I43iQe

Answer 2

So this is my process for solving it according to what you showed.
f(x) = sec(x)
x=44
a=45
L(x)=f(a) + f'(a)(x-a)
f(a)= sec(45) = 2/sqrt(2)
f'(x) = secxtanx
f'(45) = 2/sqrt(2)
L(x) = 2/sqrt(2) + [2/sqrt(2)](44-45)

Answer 3

@Dean.Shyy

Answer 4

These may also help you: http://is.gd/ayWKgK & http://is.gd/wFcbM8 & http://is.gd/4mOktp

Answer 5

What am I doing wrong? Because when I followed through I always end up with 0?

Answer 6

Maybe you can enter your values into this: http://is.gd/nQhaEY

Answer 7

Okay I think I figured it out!

f(x)=sec(44)
f'(x)=secxtanx
f'(a)=2/sqrt(2)
f(a)=2/sqrt(2)
L(x)=2/sqrt(2) + 2/sqrt(2)[x-pi/4]
=2/sqrt(2) + 2/sqrt(2)x - pi/sqrt(2)
=2/sqrt(2) + pi/sqrt(2) - pi/sqrt(2)
=2/sqrt(2)=sec(22pi/90) APPROXIMATELY = sec(pi/4)

I checked it with my calculator and the answers are only about .02 off, and I think that's right. What do you think @Dean.Shyy?