TMUA practice: Stationary Points, problem 4

Define $\;g(x) = x^{4} + 4 x$. The equation $\;g(x) = -c\;$ has a real root iff $-c$ lies in the range of $g$.

Find the minimum of $g$. Differentiate: $g'(x) = 4 x^{3} + 4 = 4(x^{3} + 1)$. Set $g'(x) = 0$: $x = -1$. Compute $g(-1) = 1 - 4 = -3$.

As $x \to \pm\infty$, $g \to +\infty$. So $g$ has range $[-3,\, \infty)$, with minimum $-3$ at $x = -1$.

For $g(x) = -c$ to have a real root, $-c \ge -3$, i.e. $\;c \le 3$. This is option D.

Why the other options fail.

• A, B: use $c \le 2$ or $c < 2$ — the minimum of $g$ is $-3$, not $-2$.
• C, $c < 3$: drops the equality case; at $c = 3$, $g(x) = -3$ has the real solution $x = -1$.

The lesson: ‘polynomial has at least one real root’ = ‘right-hand side is in the range’. Find the global minimum (or maximum) of the polynomial.