For what values of x is the function f(x) = 2 | vedclass

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(D) To find the intervals where the function $f(x) = 2x^3 - 9x^2 + 12x + 29$ is decreasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 -

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$1 < x < 2$

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(D) To find the intervals where the function $f(x) = 2x^3 - 9x^2 + 12x + 29$ is decreasing,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2x^3 - 9x^2 + 12x + 29) = 6x^2 - 18x + 12$.For the function to be decreasing,we must have $f'(x) $6x^2 - 18x + 12 Dividing by $6$,we get $x^2 - 3x + 2 Factoring the quadratic expression,we get $(x - 1)(x - 2) The inequality $(x - 1)(x - 2) Therefore,the function is decreasing for $1 < x < 2$.

For what values of $x$ is the function $f(x) = 2x^3 - 9x^2 + 12x + 29$ decreasing?

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