The interval of the decreasing function f(x) | vedclass

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(B) Given $f(x) = x^3 - x^2 - x - 4$. To find the interval where the function is decreasing,we find the derivative $f'(x)$ and set it to be less than

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$\left( -\frac{1}{3}, 1 \right)$

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(B) Given $f(x) = x^3 - x^2 - x - 4$. To find the interval where the function is decreasing,we find the derivative $f'(x)$ and set it to be less than zero: $f'(x) = 3x^2 - 2x - 1$. For the function to be decreasing,$f'(x) $3x^2 - 2x - 1 Factorizing the quadratic expression: $3x^2 - 3x + x - 1 $3x(x - 1) + 1(x - 1) $(3x + 1)(x - 1) For the product to be negative,the factors must have opposite signs. Case $1$: $3x + 1 > 0$ and $x - 1 $x > -\frac{1}{3}$ and $x Thus,the interval is $x \in \left( -\frac{1}{3}, 1 \right)$.

The interval of the decreasing function $f(x) = x^3 - x^2 - x - 4$ is

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