We are given a cubic function with a condition on its coefficients, and we need to determine whether it has any local extrema (maxima or minima).\n\nGiven:\n- $f(x) = x^3 + bx^2 + cx + d$\n- Condition: $0 < b^2 < c$\n\nWe need to find: whether $f$ has any extremum.\n\n**Strategy:** For a cubic to have extrema, its first derivative (a quadratic) must have two distinct real roots. We will check the discriminant of $f'(x)$ and use the given condition $b^2 < c$ to determine if real roots exist.\n\n**Step 1: Find $f'(x)$**\n\nUsing the power rule:\n\n$f'(x) = 3x^2 + 2bx + c$\n\nThis is a quadratic in $x$ with coefficients $A = 3$, $B = 2b$, $C = c$.\n\n**Step 2: Compute the discriminant of $f'(x)$**\n\nUsing the discriminant formula $D = B^2 - 4AC$:\n\n$D = (2b)^2 - 4(3)(c)$\n\n$= 4b^2 - 12c$\n\n$= 4(b^2 - 3c)$\n\n**Step 3: Use the given condition to determine the sign of $D$**\n\nWe are given $b^2 < c$. Since $c > 0$ (because $b^2 > 0$ and $b^2 < c$), we have $c < 3c$, so:\n\n$b^2 < c < 3c$\n\n$b^2 - 3c < 0$\n\n$D = 4(b^2 - 3c) < 0$\n\nThis means: $f'(x) = 0$ has no real roots. The quadratic $f'(x) = 3x^2 + 2bx + c$ never touches zero.\n\n**Step 4: Determine the sign of $f'(x)$**\n\nSince the leading coefficient of $f'(x)$ is $3 > 0$ and the quadratic has no real roots, $f'(x) > 0$ for all $x \in \mathbb{R}$.\n\nThis means: $f$ is strictly increasing on all of $\mathbb{R}$, so it has no local maximum or minimum.\n\n**Verification:** Consider a specific case: $b = 1$, $c = 2$ (satisfying $0 < 1 < 2$). Then $f'(x) = 3x^2 + 2x + 2$. Its discriminant is $4 - 24 = -20 < 0$, confirming no real roots. $f'(0) = 2 > 0$, so $f'(x) > 0$ always. Checks out.\n\n**Summary:** We used the discriminant of the derivative to check for critical points. The key insight is that $b^2 < c$ forces $b^2 < 3c$, making the discriminant negative. Since $f'(x) > 0$ everywhere, $f$ has no extremum.

Application of Derivatives Concepts