Let f: R rightarrow R be defined as f(x) = be | vedclass

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(A) For $x > 0$,$f'(x) = -4x^2 + 4x + 3$. Setting $f'(x) > 0$,we have $-4x^2 + 4x + 3 > 0$,which implies $4x^2 - 4x - 3 < 0$. Factoring gives $(2x - 3

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$\left(-1, \frac{3}{2}\right)$

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(A) For $x > 0$,$f'(x) = -4x^2 + 4x + 3$. Setting $f'(x) > 0$,we have $-4x^2 + 4x + 3 > 0$,which implies $4x^2 - 4x - 3 Factoring gives $(2x - 3)(2x + 1) Since $x > 0$,$f(x)$ is increasing in $\left(0, \frac{3}{2}\right)$. For $x \leq 0$,$f'(x) = 3e^x + 3xe^x = 3e^x(1 + x)$. Setting $f'(x) > 0$,since $3e^x > 0$ for all $x$,we need $1 + x > 0$,which means $x > -1$. Thus,for $x \leq 0$,$f(x)$ is increasing in $(-1, 0]$. Combining both intervals,$f(x)$ is increasing in $(-1, 0] \cup (0, \frac{3}{2}) = \left(-1, \frac{3}{2}\right)$.

Let $f: R \rightarrow R$ be defined as $f(x) = \begin{cases} -\frac{4}{3}x^3 + 2x^2 + 3x, & x > 0 \\ 3xe^x, & x \leq 0 \end{cases}$. Then $f$ is an increasing function in the interval:

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