How do you find the linearization of $f(x) = x^(1/2)$ at x=25?

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Answer 1 · 2016-11-23T19:33:29

The linearization is the tangent line. (Or maybe it is more helpful to say: it is a way of thinking about and using the tangent line.)

$f(x) = x^(1/2) = sqrtx$

At $x=25$ , we have $y = f(25) = 5$

$f'(x) = 1/(2sqrtx)$ so at $x=25$ , the slope of the tangent line is $m=f'(25) = 1/10$

Equation of tangent line in point-slope form:

$y-5 = 1/10(x-25)$

Linearization at $a=1$ (in a form I am used to):

$L(x) = 5+1/10(x-25)$

Free example of using the linearization

$sqrt28 = f(28)$

$f(28) ~~ L(28) = 5+1/10(28-25) = 5+1/10(3) = 5+0.3 = 5.3$

(The point on the tangent line with $x$ -coordinate $28$ has $y$ -coordinate $5.3$ .)

How do you find the linearization of f(x) = x^(1/2) f(x) = x^(1/2) at x=25?