Question

How do you use linear approximation about x=100 to estimate `1/sqrt(99.8)`?

Answer 1

`f(x) = 1/sqrtx`
`f(x) = x^(-1/2)`

`f(100) = 1/sqrt100 = 1/10 = .1`

`f'(x) = (-1/2)(x^(-3/2))`
`f'(100) = (-1/2)(100)^(-3/2) = -.0005`

`L(x) = -.0005(x-100)+.1`
`L(99.8) = -.0005(-.2)+.1 = .1001`