The function f(x)=x^3+a x^2+b x+c with a^2 le | vedclass

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Question

(B) Given the function $f(x)=x^3+a x^2+b x+c$. To find the extreme values,we differentiate $f(x)$ with respect to $x$: $f^{\prime}(x)=3 x^2+2 a x+b$.

Vedclass · Accepted Answer

no extreme value

Vedclass · Answer

(B) Given the function $f(x)=x^3+a x^2+b x+c$. To find the extreme values,we differentiate $f(x)$ with respect to $x$: $f^{\prime}(x)=3 x^2+2 a x+b$. For extreme values,we set $f^{\prime}(x)=0$: $3 x^2+2 a x+b=0$. The roots of this quadratic equation are given by the quadratic formula: $x = \frac{-2 a \pm \sqrt{(2 a)^2 - 4(3)(b)}}{2(3)} = \frac{-2 a \pm \sqrt{4 a^2 - 12 b}}{6} = \frac{-2 a \pm 2 \sqrt{a^2 - 3 b}}{6} = \frac{-a \pm \sqrt{a^2 - 3 b}}{3}$. Given the condition $a^2 \leq 3 b$,the term $a^2 - 3 b \leq 0$. If $a^2 - 3 b If $a^2 = 3 b$,then $f^{\prime}(x) = 3(x + \frac{a}{3})^2$,which is always $\geq 0$,so the function is monotonically increasing and has no local extrema. Thus,the function has no extreme value.

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

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