Animated Solution for Mathematics - Differentiation: Let $f(x)=\{\begin{array}{ll}x{e}^{ax},& x\le 0\\ x+a{x}^2-x^3,& x>0\end{array}$. Where $a$ is a positive constant. Find the interval in which $f^{\prime }(x)$ is increasing.

We are given the piecewise function:

Our objective is to find the interval where $f^{\prime }(x)$ is increasing. A function is increasing if its derivative is positive; therefore, we require the second derivative to be strictly positive, i.e., $f^{\prime \prime }(x)>0$.

The Left Domain ($x\le 0$)

In the region $x\le 0$, the function is defined as $f(x)=x{e}^{ax}$. Applying the product rule, the first derivative is:

Differentiating once more to find the second derivative:

Factoring out $a{e}^{ax}$, we obtain:

Since $a>0$ and $e^{ax}>0$ for all $x$, the sign of $f^{\prime \prime }(x)$ depends solely on the term $(2+ax)$. Setting $2+ax>0$ yields $x>-\frac{2}{a}$. Given the domain constraint $x\le 0$, the valid interval is $(-\frac{2}{a},0]$.

The Right Domain ($x>0$)

For $x>0$, the function is a polynomial: $f(x)=x+a{x}^2-x^3$. The first derivative is:

Differentiating again, we find:

We require $f^{\prime \prime }(x)>0$, which implies $2a-6x>0$, or $x<\frac{a}{3}$. Combining this with the domain $x>0$, we obtain the interval $(0,\frac{a}{3})$.

The Boundary and Conclusion

We must verify the behavior at the junction $x=0$. We check the limit of $f^{\prime \prime }(x)$ from both sides:

Since $a>0$, the second derivative is continuous and positive at $x=0$. We can therefore merge the intervals $(-\frac{2}{a},0]$, $\{0\}$, and $(0,\frac{a}{3})$ into a single continuous interval.

The final interval where $f^{\prime }(x)$ is increasing is: