Question

How do you find the linearization at a=16 of `f(x) = x¾`?

Answer 1

It is the equation of the tangent line at `(16,f(16))`

Explanation:

`f(x) = x¾`

`f(16) - 16^(3/4) = 8`

`f'(x) = 3/4 x^(-1/4)`, so `f'(16) = 3/8`

The tangent line has point-slope equation
`y-8 = 3/8(x-16)`,

so the linearization is

`L(x) = 8 + 3/8(x-16)`