# Linear and quadratic approximation. a. Find the linear approximating polynomial for the following functions centered at the given point a.

a. Find the linear approximating polynomial for the following functions centered at the given point a.
b. Find the quadratic approximating polynomial for the following functions centered at a.
c. Use the polynomials obtained in parts (a) and (b) to approximate the given quantity.
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{\frac{{1}}{{3}}}},{a}={8}$$ approximiate $$\displaystyle{7.5}^{{\frac{{1}}{{3}}}}$$

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Margot Mill
Given function:
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{\frac{{1}}{{3}}}},{a}={8}$$
a) The linear approximating polynomial for the functions is given by
$$\displaystyle{p}_{{1}}{\left({x}\right)}={f{{\left({a}\right)}}}+{f}'{\left({a}\right)}{\left({x}-{a}\right)}$$
$$\displaystyle{p}_{{1}}{\left({x}\right)}={8}^{{\frac{{1}}{{3}}}}+{\frac{{{1}}}{{{3}}}}{8}^{{-\frac{{2}}{{3}}}}{\left({x}-{a}\right)}$$
$$\displaystyle{p}_{{1}}{\left({x}\right)}={2}+{\frac{{{1}}}{{{2}}}}{\left({x}-{a}\right)}$$
b) The quadratic approximating polynomial for the function is given by
$$\displaystyle{p}_{{2}}{\left({x}\right)}={f{{\left({a}\right)}}}+{\frac{{{1}}}{{{2}}}}{f}{''}{\left({a}\right)}{\left({x}-{8}\right)}^{{2}}$$
$$\displaystyle{p}_{{2}}{\left({x}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({x}-{8}\right)}+{\frac{{{1}}}{{{2}}}}\times{\frac{{-{2}}}{{{9}}}}\times{8}^{{-\frac{{5}}{{3}}}}{\left({x}-{8}\right)}^{{2}}$$
$$\displaystyle{p}_{{2}}{\left({x}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({x}-{8}\right)}-{\frac{{{1}}}{{{288}}}}{\left({x}-{8}\right)}^{{2}}$$
c) The polynomials obtained in parts (a) and (b) to approximate the given quantity.
$$\displaystyle{p}_{{1}}{\left({x}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({x}-{8}\right)}$$
Substitute the value of $$\displaystyle{x}={7.5}$$
$$\displaystyle{p}_{{1}}{\left({7.5}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({7.5}-{8}\right)}$$
$$\displaystyle{p}_{{1}}{\left({7.5}\right)}={1.9583}$$
$$\displaystyle{p}_{{2}}{\left({x}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({x}-{8}\right)}-{\frac{{{1}}}{{{288}}}}{\left({x}-{8}\right)}^{{2}}$$
Substitute the value of x=7.5
$$\displaystyle{p}_{{2}}{\left({7.5}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({7.5}-{8}\right)}-{\frac{{{1}}}{{{288}}}}{\left({7.5}-{8}\right)}^{{2}}$$
$$\displaystyle{p}_{{2}}{\left({7.5}\right)}={1.95747}$$