Animated Solution for Mathematics - Differentiation: The function $f(x)=2|x|+|x+2|-||x+2|-2|x||$ has a local minimum or a local maximum at $x=$

Imagine you are staring at a function that looks like a tangled knot of absolute values:

At first glance, it feels like a trap. You might be tempted to start splitting this into five or six different intervals, calculating the slope for each, and drowning in a sea of algebraic signs.

But stop. Take a deep breath. In JEE Advanced, the most complex-looking expressions often hide a beautiful, elegant secret.

Let us perform a substitution to peel back the layers. Let $A=|x+2|$ and $B=2|x|$.

Suddenly, our function transforms into $f(x)=A+B-|A-B|$. Does this look familiar? It is the classic algebraic identity for the minimum of two numbers:

By multiplying by $2$, we see that $2\min (A,B)=A+B-|A-B|$. Just like that, the 'knot' unties itself.

Our function is simply $f(x)=2\min (|x+2|,2|x|)$. We are no longer dealing with a complex expression; we are simply looking for the 'lower envelope' of two V-shaped graphs.

Let's define our two components: $y_1=|x+2|$ and $y_2=2|x|$.

$y_1=|x+2|$ is a standard V-shape with its vertex at $x=-2$. It represents a steady, symmetric growth.

$y_2=2|x|$ is a much steeper V-shape with its vertex at $x=0$.

Our goal is to trace the minimum of these two. To find where the 'leadership' switches from one graph to the other, we must find their intersection points by setting $|x+2|=2|x|$.

This gives us two scenarios: 1. $x+2=2x$, which yields $x=2$. 2. $x+2=-2x$, which yields $3x=-2$, or $x=-\frac{2}{3}$.

These points, $x=-\frac{2}{3}$ and $x=2$, are the 'crossover' points where the function switches its identity.

Now, let's walk along the x-axis: - For $x\le -\frac{2}{3}$, the graph $y_1$ is below $y_2$. So, $f(x)=2|x+2|$. - Between $-\frac{2}{3}$ and $2$, the graph $y_2$ dips below $y_1$. So, $f(x)=2(2|x|)=4|x|$. - For $x\ge 2$, $y_1$ returns to being the lower graph. So, $f(x)=2|x+2|$.

Finding Extrema

Now, we look for the peaks and valleys: - At $x=-2$, we have a sharp valley in $y_1$. Since $y_1$ is the minimum here, $f(x)$ also has a local minimum. - At $x=0$, we have a sharp valley in $y_2$. Since $y_2$ is the minimum here, $f(x)$ has another local minimum. - At $x=-\frac{2}{3}$, the function transitions from the $y_1$ branch to the $y_2$ branch. If you visualize the graph, you will see it climbs up to a peak before dropping down to the next valley. This is a local maximum.

What about $x=2$? At this point, the slope changes, but the function continues to increase or decrease in a way that doesn't form a peak or a valley. However, in the context of identifying critical points where the function's behavior changes, $x=2$ is a vital transition point.

By breaking down the complexity, we didn't just solve a problem; we mapped the landscape of the function. You have successfully navigated the absolute value maze. Remember, in physics and mathematics, the most daunting problems are often just simple concepts wearing a mask.