Let f(x) = x^3 + bx^2 + cx + d where 0

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(D) Given the function $f(x) = x^3 + bx^2 + cx + d$. First,we find the derivative of the function: $f'(x) = 3x^2 + 2bx + c$. To determine the nature o

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Is strictly increasing

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(D) Given the function $f(x) = x^3 + bx^2 + cx + d$. First,we find the derivative of the function: $f'(x) = 3x^2 + 2bx + c$. To determine the nature of the function,we examine the discriminant $D$ of the quadratic expression $f'(x)$: $D = (2b)^2 - 4(3)(c) = 4b^2 - 12c$. We are given that $0 Since $b^2 Therefore,$D = 4b^2 - 12c Since $b^2 > 0$,$c$ must also be greater than $0$ (because $c > b^2$). Thus,$D Since the discriminant $D  0$ for all $x \in \mathbb{R}$. Since the derivative $f'(x)$ is strictly positive for all real $x$,the function $f(x)$ is strictly increasing.

Let $f(x) = x^3 + bx^2 + cx + d$ where $0 < b^2 < c$. Then $f(x)$:

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