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

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Answer 1 · 2015-11-06T22:51:28

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

$f(x) = x¾$

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

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

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

so the linearization is

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

How do you find the linearization at a=16 of f(x) = x¾f(x) = x¾?