If f(x)=2x^3-15x^2-144x-7,then f(x) is strict | vedclass

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(B) Given the function $f(x)=2x^3-15x^2-144x-7$. To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$: $f'(x) = \f

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$(-3,8)$

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(B) Given the function $f(x)=2x^3-15x^2-144x-7$. To find the intervals where $f(x)$ is strictly decreasing,we find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(2x^3-15x^2-144x-7) = 6x^2-30x-144$. For $f(x)$ to be strictly decreasing,$f'(x) $6x^2-30x-144 Dividing by $6$,we get $x^2-5x-24 Factoring the quadratic expression: $(x-8)(x+3) The roots are $x=8$ and $x=-3$. The inequality holds for $x$ in the interval $(-3, 8)$. Thus,$f(x)$ is strictly decreasing in $(-3, 8)$.

If $f(x)=2x^3-15x^2-144x-7$,then $f(x)$ is strictly decreasing in

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