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

\(\displaystyle{p}_{{2}}{\left({7.5}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({7.5}-{8}\right)}-{\frac{{{1}}}{{{288}}}}{\left({7.5}-{8}\right)}^{{2}}\)

\(\displaystyle{p}_{{2}}{\left({7.5}\right)}={1.95747}\)

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{\frac{{1}}{{3}}}},{a}={8}\)

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

\(\displaystyle{p}_{{2}}{\left({7.5}\right)}={2}+{\frac{{{1}}}{{{12}}}}{\left({7.5}-{8}\right)}-{\frac{{{1}}}{{{288}}}}{\left({7.5}-{8}\right)}^{{2}}\)

\(\displaystyle{p}_{{2}}{\left({7.5}\right)}={1.95747}\)