If f(x) = x^3 - 6x^2 + 9x + 3 is a decreasing | vedclass

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(B) Given the function $f(x) = x^3 - 6x^2 + 9x + 3$. To find the interval where the function is decreasing,we find the derivative $f'(x)$. $f'(x) = \f

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$(1, 3)$

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(B) Given the function $f(x) = x^3 - 6x^2 + 9x + 3$. To find the interval where the function is decreasing,we find the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x + 3) = 3x^2 - 12x + 9$. $A$ function is decreasing when $f'(x) $3x^2 - 12x + 9 Dividing by $3$,we get $x^2 - 4x + 3 Factoring the quadratic,we get $(x - 1)(x - 3) The roots of the equation are $x = 1$ and $x = 3$. The inequality $(x - 1)(x - 3) Therefore,$x \in (1, 3)$.

If $f(x) = x^3 - 6x^2 + 9x + 3$ is a decreasing function,then $x$ lies in

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