Animated Solution for Mathematics - Differentiation: The sum of all local minimum values of the function $f(x)=\{\begin{array}{ll}1-2x,& x<-1\\ \frac{1}{3}(7+2|x|),& -1\le x\le 2\\ \frac{1}{18}(x-4)(x-5),& x>2\end{array}$ is

Imagine you are standing on a vast, rugged terrain. This terrain isn't a single, smooth hill; it is a landscape composed of three distinct, connected sections.

This is exactly what a piecewise function like $f(x)$ represents. To find the local minima, we aren't just solving an equation; we are exploring this terrain to find the lowest valleys.

The Linear Descent ($x<-1$)

In the first region, where $x<-1$, the function is defined as $f(x)=1-2x$. This is a simple, straight path with a constant negative slope of $-2$.

As you walk along this path, you are constantly descending. There is no valley here, just a continuous slide downward. Since the function is strictly decreasing, we can confidently say there are no local minima in this region.

The Modulus Valley ($x=0$)

Now, we enter the second region, defined for $-1\le x\le 2$ as:

This is where the terrain gets interesting. The modulus function $|x|$ acts like a mirror, reflecting the path. We must split this at $x=0$.

For $-1\le x<0$, the function is $f(x)=\frac{1}{3}(7-2x)$, which continues our descent. But at $x=0$, the path hits a sharp turn.

For $0\le x\le 2$, the function becomes $f(x)=\frac{1}{3}(7+2x)$, and suddenly, the slope becomes positive. We have found our first valley.

By substituting $x=0$ into the function, we find the first local minimum value:

The Jump Discontinuity ($x=2$)

As we approach $x=2$, we encounter a cliff. If we look at the left-hand limit, $f(2^{-})=\frac{11}{3}$, but the right-hand limit, using the third definition, drops to:

This is a massive jump discontinuity. Because the value at $x=2$ is higher than the values immediately to its right, this point is a local maximum, not a minimum. We must be careful not to fall into this trap!

The Parabolic Vertex ($x=4.5$)

Finally, we reach the third region, where $x>2$. The function is $f(x)=\frac{1}{18}(x-4)(x-5)$.

If we expand this, we get $f(x)=\frac{1}{18}(x^2-9x+20)$, which is an upward-opening parabola. Every parabola has a vertex—the absolute lowest point of that curve.

We find this vertex using the formula $x=-\frac{b}{2a}$, which gives us $x=\frac{9}{2}=4.5$. This is well within our region $x>2$.

Let's calculate the value at this valley:

This is our second local minimum!

We have traversed the entire landscape and identified our two valleys. The first valley is at $f(0)=\frac{7}{3}$, and the second is at $f(\frac{9}{2})=-\frac{1}{72}$.

The problem asks for the sum of all local minimum values. Let's combine them:

To add these, we need a common denominator, which is $72$. Converting $\frac{7}{3}$, we get $\frac{168}{72}$.

Now, $\frac{168}{72}-\frac{1}{72}=\frac{167}{72}$. We have successfully navigated the terrain and found the sum of our local minima. The final result is $\frac{167}{72}$.