MCQ
Let $f(x) = {x^2} + 4x + 1$. Then
- A$f(x) > 0$ for all $x$
- B$f(x) > 1$ when $x \ge 0$
- ✓$f(x) \ge 1$ when $x \le - 4$
- D$f(x) = f( - x)$ for all $x$
$f(x) \ge 1 \Rightarrow {x^2} + 4x + 1 \ge 1\, \Rightarrow {x^2} + 4x \ge 0$
==> $x \le - 4$or $x \ge 0$.
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