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

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(C) Given the function $f(x) = [x(x - 3)]^2 = (x^2 - 3x)^2$. To find where the function is increasing,we calculate the derivative $f'(x)$ and set $f'(

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$0 < x < 3/2$

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(C) Given the function $f(x) = [x(x - 3)]^2 = (x^2 - 3x)^2$. To find where the function is increasing,we calculate the derivative $f'(x)$ and set $f'(x) > 0$. $f'(x) = 2(x^2 - 3x) \cdot (2x - 3) = 2x(x - 3)(2x - 3)$. Setting $f'(x) > 0$,we get $2x(x - 3)(2x - 3) > 0$,which simplifies to $x(x - 3)(2x - 3) > 0$. The critical points are $x = 0$,$x = 3/2$,and $x = 3$. Using the wavy curve method (sign scheme) on the number line: For $x \in (0, 3/2)$,$f'(x) > 0$. For $x \in (3/2, 3)$,$f'(x) For $x \in (3, \infty)$,$f'(x) > 0$. Thus,the function is increasing for $x \in (0, 3/2) \cup (3, \infty)$.

For what values of $x$ is the function $f(x) = [x(x - 3)]^2$ an increasing function?

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