Find the local maximum and local minimum values for the function given by $f(x) = x^{2}$.

(N/A) $f(x) = x^{2}$
$\therefore f'(x) = 2x$
Now,set $f'(x) = 0$ to find the critical points:
$2x = 0 \Rightarrow x = 0$
Thus,$x = 0$ is the only critical point.
We calculate the second derivative: $f''(x) = 2$.
Since $f''(0) = 2 > 0$,by the second derivative test,$x = 0$ is a point of local minima.
The local minimum value of $f$ at $x = 0$ is $f(0) = (0)^{2} = 0$.
There is no local maximum value for this function as $f(x) \to \infty$ as $x \to \pm \infty$.