Animated Solution for Mathematics - Limits, Continuity and Differentiability: The function $f(x)=|x^2-2x-3|\cdot e^{|9x^2-12x+4|}$ is not differentiable at exactly:

The function provided is $f(x)=|x^2-2x-3|\cdot e^{|9x^2-12x+4|}$. To determine its points of non-differentiability, we decompose it into two distinct components: $h(x)=|x^2-2x-3|$ and $g(x)=e^{|9x^2-12x+4|}$.

We treat $h(x)$ as the potential source of "sharp corners" and $g(x)$ as a smooth, continuous operator. Our objective is to identify where the product $f(x)=h(x)\cdot g(x)$ fails to be differentiable.

We first examine $h(x)=|x^2-2x-3|$. The absolute value function creates sharp cusps at the roots of the internal quadratic expression.

Setting the quadratic to zero, we solve:

Factoring the expression, we obtain:

This yields roots at $x=-1$ and $x=3$. At these specific points, the graph of the parabola crosses the $x$-axis, and the absolute value operation reflects the negative portion, creating sharp corners. Consequently, $h(x)$ is non-differentiable at $x=-1$ and $x=3$.

Next, we analyze $g(x)=e^{|9x^2-12x+4|}$. While the absolute value might initially suggest further non-differentiability, we observe that the internal expression is a perfect square:

Since $(3x-2{)}^2\ge 0$ for all real $x$, the absolute value is redundant. We can simplify the function to:

Because $g(x)$ is a composition of an exponential function and a polynomial, it is smooth and differentiable for all $x\in \mathbb{R}$. There are no sharp corners introduced by this component.

We now consider the product $f(x)=h(x)\cdot g(x)$. A product of a non-differentiable function and a smooth function remains non-differentiable at the original points of non-differentiability, unless the smooth function vanishes at those points.

We check if $g(x)$ can "flatten" the corners by evaluating it at the critical points:

Since $g(x)=e^{(3x-2{)}^2}$ is an exponential function, it is strictly positive for all real $x$ and never equals zero. Therefore, the sharp corners of $h(x)$ are not smoothed out by $g(x)$.

Because the smooth component $g(x)$ never vanishes at the critical points, the sharp corners at $x=-1$ and $x=3$ persist in the final function $f(x)$.

The function $f(x)$ is non-differentiable at exactly two points.