The given function is \(\displaystyle f{{\left({x}\right)}}{i}{s}{\left({x}+{9}\right)}^{2}.\)

Obtain the first and second derivative of the function as follows.

\(\displaystyle f{{\left({x}\right)}}={\left({x}+{9}\right)}^{2}\)

\(\displaystyle{f}'{\left({x}\right)}={2}{\left({x}+{9}\right)}\)

\(\displaystyle{f}'{\left({x}\right)}={2}\)

Find the critical points as follows

\(\displaystyle{f}'{\left({x}\right)}={0}\)

\(\displaystyle{2}{\left({x}+{9}\right)}={0}\)

\(\displaystyle\frac{{{2}{\left({x}+{9}\right)}}}{{2}}=\frac{0}{{2}}\)

\(\displaystyle{x}=-{9}\)

Thus, the function has critical point at \(\displaystyle{x}=-{9}\)

Since the second derivative is a constant function and is greater than zero for all values of x, the function has only local minima.

Thus, the function has local minima or minimum at \(\displaystyle{\left(–{9},{0}\right)}.\)

Obtain the first and second derivative of the function as follows.

\(\displaystyle f{{\left({x}\right)}}={\left({x}+{9}\right)}^{2}\)

\(\displaystyle{f}'{\left({x}\right)}={2}{\left({x}+{9}\right)}\)

\(\displaystyle{f}'{\left({x}\right)}={2}\)

Find the critical points as follows

\(\displaystyle{f}'{\left({x}\right)}={0}\)

\(\displaystyle{2}{\left({x}+{9}\right)}={0}\)

\(\displaystyle\frac{{{2}{\left({x}+{9}\right)}}}{{2}}=\frac{0}{{2}}\)

\(\displaystyle{x}=-{9}\)

Thus, the function has critical point at \(\displaystyle{x}=-{9}\)

Since the second derivative is a constant function and is greater than zero for all values of x, the function has only local minima.

Thus, the function has local minima or minimum at \(\displaystyle{\left(–{9},{0}\right)}.\)