How do you find the linearization of $f(x) = x^(1/2)$ at a=16?

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Answer 1 · 2017-02-05T17:10:02

The linearization is:

$y=x/8+2$

We can approximate a curve using its tangent line:

$y = f(a)+f'(a)(x-a)$

For $a=16$ we have:

$f(a) = 16^(1/2)=sqrt(16) = 4$

$f'(x) = d/dx x^(1/2) = 1/(2sqrtx)$

$f'(a) = 1/(2sqrt16) = 1/8$

so the linarization is:

$y=4+1/8(x-16)$

$y=x/8+2$

How do you find the linearization of f(x) = x^(1/2)f(x) = x^(1/2) at a=16?