The condition for the function f(x) = x^3 + p | vedclass

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Question

(A) For a function $f(x)$ to have no extreme value, its derivative $f'(x)$ must not change sign. This means $f'(x)$ must be either always non-negative

Vedclass · Accepted Answer

$p^2 < 3q$

Vedclass · Answer

(A) For a function $f(x)$ to have no extreme value, its derivative $f'(x)$ must not change sign. This means $f'(x)$ must be either always non-negative or always non-positive. Given $f(x) = x^3 + px^2 + qx + r$. The derivative is $f'(x) = 3x^2 + 2px + q$. For $f'(x)$ to have no sign change, the quadratic equation $3x^2 + 2px + q = 0$ must have no real roots or a repeated root. This implies the discriminant $D \le 0$. The discriminant $D = (2p)^2 - 4(3)(q) = 4p^2 - 12q$. Setting $D \le 0$, we get $4p^2 - 12q \le 0$. Dividing by $4$, we get $p^2 - 3q \le 0$, or $p^2 \le 3q$. Given the options, the condition $p^2 < 3q$ is the correct choice.

The condition for the function $f(x) = x^3 + px^2 + qx + r$ $(x \in R)$ to have no extreme value is:

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