Let f(x) = begin{cases} x e^{3x}, & x le 0 2 | vedclass

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(D) First,we find $f'(x)$ for both intervals:For $x \le 0$,$f(x) = x e^{3x}$. Using the product rule,$f'(x) = e^{3x} + 3x e^{3x} = e^{3x}(1 + 3x)$.For

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(D) First,we find $f'(x)$ for both intervals:For $x \le 0$,$f(x) = x e^{3x}$. Using the product rule,$f'(x) = e^{3x} + 3x e^{3x} = e^{3x}(1 + 3x)$.For $x > 0$,$f(x) = 2x^3 + x$. Then $f'(x) = 6x^2 + 1$.Thus,$f'(x) = \begin{cases} e^{3x}(1 + 3x), & x \le 0 \ 6x^2 + 1, & x > 0 \end{cases}$.For $f'(x)$ to be an increasing function,its derivative $f''(x)$ must be greater than $0$.For $x Setting $f''(x) > 0$,we get $3e^{3x}(2 + 3x) > 0$. Since $3e^{3x} > 0$ for all $x$,we need $2 + 3x > 0$,which implies $x > -\frac{2}{3}$.For $x > 0$,$f''(x) = \frac{d}{dx}[6x^2 + 1] = 12x$. Setting $f''(x) > 0$,we get $12x > 0$,which implies $x > 0$.Combining these intervals for $x  0$,we get $x \in (-\frac{2}{3}, 0) \cup (0, \infty)$.Checking continuity of $f'(x)$ at $x=0$: $\lim_{x 	o 0^-} f'(x) = e^0(1+0) = 1$ and $\lim_{x 	o 0^+} f'(x) = 6(0)^2 + 1 = 1$. Since $f'(0) = 1$,$f'(x)$ is continuous at $x=0$.Thus,$f'(x)$ is increasing for $x \in (-\frac{2}{3}, \infty)$.

Let $f(x) = \begin{cases} x e^{3x}, & x \le 0 \\ 2x^3 + x, & x > 0 \end{cases}$. Find the complete set of values of $x$ for which $f'(x)$ is an increasing function.

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