How do you find the linear approximation of the function $g (x) = \sqrt[5]{1 + x}$ $g(x)=root5(1+x)$ at a=0?

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Answer 1 · 2016-12-30T13:22:29

$\sqrt[5]{1 + x} ≅ 1 + \frac{x}{5}$ $root(5)(1+x) ~= 1+x/5$

The best linear approximation of $g (x)$ $g(x)$ around $x = 0$ $x=0$ is its tangent line:

$y(x) = g(0)+g'(0)x$

Given that:

$g(x) = root(5)(1+x) = (1+x)^(1/5)$

$g'(x) = 1/5(1+x)^(-4/5)$

we have:

$g(0) = 1$

$g'(0) = 1/5$

so:

$root(5)(1+x) ~= 1+x/5$

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How do you find the linear approximation of the function g(x)=root5(1+x)g(x)=5√1+xg(x)=root5(1+x) at a=0?