Let f(x) = x^2 + 4x + 1. Then | vedclass

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(C) Given $f(x) = x^2 + 4x + 1$.For option $A$: The discriminant $D = b^2 - 4ac = 4^2 - 4(1)(1) = 16 - 4 = 12 > 0$. Since $D > 0$,the quadratic has tw

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$f(x) \ge 1$ when $x \le -4$

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(C) Given $f(x) = x^2 + 4x + 1$.For option $A$: The discriminant $D = b^2 - 4ac = 4^2 - 4(1)(1) = 16 - 4 = 12 > 0$. Since $D > 0$,the quadratic has two distinct real roots,so $f(x)$ takes both positive and negative values. Thus,$f(x) > 0$ is not true for all $x$.For option $B$: If $x \ge 0$,then $x^2 \ge 0$ and $4x \ge 0$. Thus $f(x) = x^2 + 4x + 1 \ge 1$. Specifically,$f(x) > 1$ for $x > 0$. This is a correct statement.For option $C$: $f(x) \ge 1 \implies x^2 + 4x + 1 \ge 1 \implies x^2 + 4x \ge 0 \implies x(x + 4) \ge 0$. This inequality holds when $x \le -4$ or $x \ge 0$. Thus,$f(x) \ge 1$ when $x \le -4$ is a correct statement.Note: Both $B$ and $C$ are mathematically correct statements based on the inequality $x(x+4) \ge 0$.

Let $f(x) = x^2 + 4x + 1$. Then

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