The interval for which the given function f(x | vedclass

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(A) Given function: $f(x) = 2x^3 - 3x^2 - 36x + 7$Step $1$: Find the derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36$

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$(-2, 3)$

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(A) Given function: $f(x) = 2x^3 - 3x^2 - 36x + 7$Step $1$: Find the derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36$Step $2$: For the function to be decreasing,we must have $f'(x) $6x^2 - 6x - 36 Step $3$: Divide by $6$ and factorize the quadratic expression.$x^2 - x - 6 $(x - 3)(x + 2) Step $4$: Determine the interval where the product is negative.The roots are $x = 3$ and $x = -2$. The expression $(x - 3)(x + 2)$ is negative between the roots.Thus,$-2 Therefore,the interval is $(-2, 3)$.

The interval for which the given function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is decreasing,is

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