The function f(x) = x^3 + ax^2 + bx + c where | vedclass

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Question

(B) Given the function $f(x) = x^3 + ax^2 + bx + c$. To find the extreme values, we differentiate $f(x)$ with respect to $x$: $f'(x) = 3x^2 + 2ax + b$

Vedclass · Accepted Answer

no extreme value

Vedclass · Answer

(B) Given the function $f(x) = x^3 + ax^2 + bx + c$. To find the extreme values, we differentiate $f(x)$ with respect to $x$: $f'(x) = 3x^2 + 2ax + b$. For extreme values, we set $f'(x) = 0$: $3x^2 + 2ax + b = 0$. The discriminant of this quadratic equation is $D = (2a)^2 - 4(3)(b) = 4a^2 - 12b = 4(a^2 - 3b)$. Given the condition $a^2 \leq 3b$, it follows that $a^2 - 3b \leq 0$, which implies $D \leq 0$. If $D If $D = 0$, $f'(x) = 3(x + a/3)^2$, which is always $\geq 0$, so the function is monotonically increasing and has a point of inflection, not an extreme value. Therefore, the function has no extreme value.

The function $f(x) = x^3 + ax^2 + bx + c$ where $a^2 \leq 3b$ has:

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