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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

The sum of two nonzero numbers is $4$. The minimum value of the sum of their reciprocals is

The function $f(x) = x + \sin x$ has

The condition that $f(x) = ax^3 + bx^2 + cx + d$ has no extreme value is

At $x=0$,$f(x)=\cos x-1+\frac{x^2}{2}-\frac{x^3}{3}$

The sum of the maximum and the minimum values of $f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 4$ in the interval $(0, 2)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module