Use degree 2 Taylor Polynomial to estimate value.

Question

Given $f(x)=(x+16)^{1\over3}$, use Taylor's polynomial of degree two about $c= -15$ to estimate the value of $(1.27)^{1\over3}$.

Can anyone please help with this question, shown steps would also be greatly appreciated.

Answer 1

The general Taylor's formula until the second order is:

$f(x)=f(x_0)+f'(x_0)(x-x_0)+{f''(x_0)\over2!}(x-x_0)^2+o(x^2)$

so for $f(x)=(x+16)^{1\over3}$ and $x_0=-15$, we have:

$$f(x)\approx(-15+16)^{1\over3}+{1\over3}(-15+16)^{-{2\over3}}(x+15)-{1\over9}(-15+16)^{-{5\over3}}(x+15)^2+o(x^2)$$ $$f(x)\approx1+{1\over3}(x+15)-{1\over9}(x+15)^2+o(x^2)$$

We want estimate $(1.27)^{1\over3}$, so we evaluate $f(x)$ for $x=-14,73$:

$$f(-14,73)=(-14,73+16)^{1\over3}=(1.27)^{1\over3}\approx1+{1\over3}(0.27)-{1\over9}(0.27)^2+o(x^2)\approx1.0819$$

Note that $(1.27)^{1\over3}=1.082932133$.