Prove that the function $h(x) = x^{3} + x^{2} + x + 1$ does not have any local maxima or local minima.

(N/A) Given the function $h(x) = x^{3} + x^{2} + x + 1$.
First,we find the derivative of the function with respect to $x$:
$h'(x) = \frac{d}{dx}(x^{3} + x^{2} + x + 1) = 3x^{2} + 2x + 1$.
To find the critical points,we set $h'(x) = 0$:
$3x^{2} + 2x + 1 = 0$.
For this quadratic equation $ax^{2} + bx + c = 0$,the discriminant $D$ is given by $D = b^{2} - 4ac$.
Here,$a = 3$,$b = 2$,and $c = 1$.
$D = (2)^{2} - 4(3)(1) = 4 - 12 = -8$.
Since the discriminant $D < 0$,the equation $3x^{2} + 2x + 1 = 0$ has no real roots.
This implies that $h'(x)$ is always positive (since the coefficient of $x^{2}$ is positive) for all $x \in \mathbb{R}$.
Since $h'(x) \neq 0$ for any $x \in \mathbb{R}$,the function $h(x)$ does not have any local maxima or local minima.