Animated Solution for Mathematics - Differentiation: Let $f:\mathbb{R}\to \mathbb{R}$ be defined as $f(x)=\{\begin{array}{ll}-\frac{4}{3}{x}^3+2x^2+3x& ,x>0\\ 3x{e}^x& ,x\le 0\end{array}$. Then $f$ is increasing function in the interval

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a problem; we are dissecting a mathematical creature with a split personality. We are looking at a function $f(x)$ defined as:

Our mission is to find where this function is increasing. In the language of calculus, this means we are hunting for the intervals where the first derivative is strictly positive, or $f^{\prime }(x)>0$.

We have two distinct mathematical landscapes to explore, and we will conquer them one by one.

The Cubic Landscape ($x>0$)

Let us stand on the right side of the y-axis, where $x>0$. Here, our function is a cubic polynomial: $f(x)=-\frac{4}{3}{x}^3+2x^2+3x$. To understand its growth, we calculate its slope using the power rule:

We require $f^{\prime }(x)>0$, so we set up the inequality $-4x^2+4x+3>0$. To simplify, we multiply by $-1$ and flip the inequality sign:

Factoring the quadratic, we look for two numbers that multiply to $-12$ and add to $-4$. This yields $(2x+1)(2x-3)<0$. The roots are $x=-\frac{1}{2}$ and $x=\frac{3}{2}$.

Using the wavy curve method, the expression is negative between the roots, giving $x\in (-\frac{1}{2},\frac{3}{2})$. Respecting our domain constraint $x>0$, we intersect these sets to find our first valid interval:

The Exponential Landscape ($x\le 0$)

Now, let us cross the boundary into the left side, where $x\le 0$. Here, the function is $f(x)=3x{e}^x$. Applying the product rule, $\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$, we get:

We need $f^{\prime }(x)>0$, so we set $3e^x(1+x)>0$. Since $e^x$ is strictly positive for all real $x$, it acts as a positive scaling factor that does not affect the inequality.

Dividing by $3e^x$, we are left with $1+x>0$, which simplifies to $x>-1$. We must intersect this with our domain constraint $x\le 0$. This yields the interval:

The Synthesis

We have conquered both territories. On the right, the function increases on $(0,\frac{3}{2})$. On the left, it increases on $(-1,0]$.

Since the function is continuous at the junction $x=0$, we can bridge these two intervals together. The union of $(-1,0]$ and $(0,\frac{3}{2})$ gives us the final, unified interval:

You have successfully navigated a piecewise function by respecting its boundaries and applying the correct tools for each domain. This is the essence of JEE Advanced mathematics—breaking down complex, intimidating problems into manageable, elegant steps.