If 27a + 9b + 3c + d = 0,then the equation 4a | vedclass

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(C) Let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. Then,$f'(x) = 4ax^3 + 3bx^2 + 2cx + d$. We are given that $f'(x) = 0$ has at least one root. Consider th

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$0$ and $3$

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(C) Let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$. Then,$f'(x) = 4ax^3 + 3bx^2 + 2cx + d$. We are given that $f'(x) = 0$ has at least one root. Consider the values of $f(x)$ at $x=0$ and $x=3$: $f(0) = a(0)^4 + b(0)^3 + c(0)^2 + d(0) + e = e$. $f(3) = a(3)^4 + b(3)^3 + c(3)^2 + d(3) + e = 81a + 27b + 9c + 3d + e$. We can rewrite $f(3)$ as $f(3) = 3(27a + 9b + 3c + d) + e$. Since it is given that $27a + 9b + 3c + d = 0$,we have $f(3) = 3(0) + e = e$. Thus,$f(0) = f(3) = e$. Since $f(x)$ is a polynomial,it is continuous on $[0, 3]$ and differentiable on $(0, 3)$. By Rolle's Theorem,there exists at least one $c \in (0, 3)$ such that $f'(c) = 0$. Therefore,the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root in the interval $(0, 3)$.

If $27a + 9b + 3c + d = 0$,then the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root between which of the following?

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