The maximum value of the function f(x) = x^3 | vedclass

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(C) Given the function $f(x) = x^3 + x^2 + x - 4$. To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 + 2x + 1$. For critica

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Does not have a maximum value

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(C) Given the function $f(x) = x^3 + x^2 + x - 4$. To find the critical points,we calculate the first derivative: $f'(x) = 3x^2 + 2x + 1$. For critical points,we set $f'(x) = 0$,which gives $3x^2 + 2x + 1 = 0$. The discriminant of this quadratic equation is $D = b^2 - 4ac = (2)^2 - 4(3)(1) = 4 - 12 = -8$. Since the discriminant $D This means the function $f(x)$ is strictly increasing for all real $x$ because the leading coefficient of the cubic polynomial is positive. Therefore,the function $f(x)$ does not have a local maximum or minimum value in the set of real numbers. Thus,the correct option is $(C)$.

The maximum value of the function $f(x) = x^3 + x^2 + x - 4$ is

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