# What is the linear approximation of g(x)=sqrt(1+x)^(1/5) at a =0?

Jan 27, 2015

(I suppose that you mean x=0)

The function, using the power properties, becomes: $y = {\left({\left(1 + x\right)}^{\frac{1}{2}}\right)}^{\frac{1}{5}} = {\left(1 + x\right)}^{\left(\frac{1}{2}\right) \left(\frac{1}{5}\right)} = {\left(1 + x\right)}^{\frac{1}{10}}$

To make a linear approximation of this function it is useful to remember the MacLaurin series, that is the Taylor's polinomial centered in zero.

This series, interrupted to the second power, is:

(1+x)^alpha=1+alpha/(1!)x+(alpha(alpha-1))/(2!)x^2...

so the linear approximation of this function is:

$g \left(x\right) = 1 + \frac{1}{10} x$