If f(x) = x^3 + bx^2 + cx + d and 0

Question

(A) Given $f(x) = x^3 + bx^2 + cx + d$. Taking the derivative with respect to $x$,we get $f'(x) = 3x^2 + 2bx + c$. For $f(x)$ to be strictly increasin

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$f(x)$ is a strictly increasing function

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(A) Given $f(x) = x^3 + bx^2 + cx + d$. Taking the derivative with respect to $x$,we get $f'(x) = 3x^2 + 2bx + c$. For $f(x)$ to be strictly increasing,we need $f'(x) > 0$ for all $x \in \mathbb{R}$. The discriminant of the quadratic $f'(x)$ is $D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$. Since $b^2 Thus,$D = 4b^2 - 12c Since $0  0$. Therefore,$D Since the leading coefficient of $f'(x)$ is $3 > 0$ and $D  0$ for all $x \in \mathbb{R}$. Hence,$f(x)$ is a strictly increasing function on $(-\infty, \infty)$.

If $f(x) = x^3 + bx^2 + cx + d$ and $0 < b^2 < c$,then in $(-\infty, \infty)$:

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