If the function f(x)=x^3+ax^2+bx+40 satisfies | vedclass

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(B) Given $f(x) = x^3 + ax^2 + bx + 40$. Since $-5$ and $4$ are roots of $f(x) = 0$, we have $f(-5) = 0$ and $f(4) = 0$. $f(-5) = (-5)^3 + a(-5)^2 + b

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$\frac{1+\sqrt{67}}{3}$

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(B) Given $f(x) = x^3 + ax^2 + bx + 40$. Since $-5$ and $4$ are roots of $f(x) = 0$, we have $f(-5) = 0$ and $f(4) = 0$. $f(-5) = (-5)^3 + a(-5)^2 + b(-5) + 40 = -125 + 25a - 5b + 40 = 25a - 5b - 85 = 0 \Rightarrow 5a - b = 17$ $(i)$. $f(4) = (4)^3 + a(4)^2 + b(4) + 40 = 64 + 16a + 4b + 40 = 16a + 4b + 104 = 0 \Rightarrow 4a + b = -26$ $(ii)$. Adding $(i)$ and $(ii)$, we get $9a = -9 \Rightarrow a = -1$. Substituting $a = -1$ in $(i)$, $5(-1) - b = 17 \Rightarrow -5 - b = 17 \Rightarrow b = -22$. Thus, $f(x) = x^3 - x^2 - 22x + 40$. By Rolle's theorem, there exists $c \in (-5, 4)$ such that $f'(c) = 0$. $f'(x) = 3x^2 - 2x - 22$. Setting $f'(c) = 0$, we get $3c^2 - 2c - 22 = 0$. Using the quadratic formula $c = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-22)}}{2(3)} = \frac{2 \pm \sqrt{4 + 264}}{6} = \frac{2 \pm \sqrt{268}}{6} = \frac{2 \pm 2\sqrt{67}}{6} = \frac{1 \pm \sqrt{67}}{3}$. Since $c \in (-5, 4)$, the value $\frac{1+\sqrt{67}}{3} \approx \frac{1+8.18}{3} \approx 3.06$ lies in the interval. Thus, one of the values of $c$ is $\frac{1+\sqrt{67}}{3}$.

If the function $f(x)=x^3+ax^2+bx+40$ satisfies the conditions of Rolle's theorem on the interval $[-5,4]$ and $-5,4$ are two roots of the equation $f(x)=0$, then one of the values of $c$ as stated in that theorem is

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