If f(x) = x^3 + ax^2 + bx + c has a minimum a | vedclass

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(C) Given $f(x) = x^3 + ax^2 + bx + c$.Find the derivative: $f'(x) = 3x^2 + 2ax + b$.Since the function has extrema at $x = 3$ and $x = -1$,$f'(x) = 0

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$a = -3, b = -9, c \in \mathbb{R}$

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(C) Given $f(x) = x^3 + ax^2 + bx + c$.Find the derivative: $f'(x) = 3x^2 + 2ax + b$.Since the function has extrema at $x = 3$ and $x = -1$,$f'(x) = 0$ at these points.Thus,the roots of $3x^2 + 2ax + b = 0$ are $x = 3$ and $x = -1$.This implies $f'(x) = k(x - 3)(x + 1) = k(x^2 - 2x - 3) = kx^2 - 2kx - 3k$.Comparing this with $3x^2 + 2ax + b$,we get $k = 3$.So,$f'(x) = 3(x^2 - 2x - 3) = 3x^2 - 6x - 9$.Comparing coefficients: $2a = -6 \implies a = -3$ and $b = -9$.Since $c$ is a constant of integration,it can be any real number,$c \in \mathbb{R}$.

If $f(x) = x^3 + ax^2 + bx + c$ has a minimum at $x = 3$ and a maximum at $x = -1$,then:

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