Animated Solution for Mathematics - Differentiation: Let $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=\{\begin{array}{ll}k-2x,& \text{if }x\le -1\\ 2x+3,& \text{if }x>-1\end{array}$. If $f$ has a local minimum at $x=-1$, then a possible value of $k$ is

Imagine you are standing on a mountain trail that suddenly splits into two distinct paths at a specific marker, $x=-1$. To your left, the path is defined by $f(x)=k-2x$, and to your right, it is $f(x)=2x+3$.

Your mission is to find the value of $k$ that turns this junction into the lowest point in the immediate vicinity—a local minimum. This is not just an algebraic exercise; it is a study of the landscape of functions.

Let us first look at the right branch, where $x>-1$. The function is $f(x)=2x+3$. This is a straight line with a positive slope of $2$.

As you walk along this path towards the junction from the right, your altitude increases. As $x$ approaches $-1$ from the right, the limit is:

Now, look at the left branch, where $x<-1$. The function is $f(x)=k-2x$. The slope here is $-2$, meaning the path is descending as you move towards the junction. This is crucial: the function is strictly decreasing as it approaches $x=-1$ from the left.

Many students panic when they see a piecewise function, immediately looking for derivatives. But stop! A local minimum is defined by the behavior of the function's values, not its slope.

A point $c$ is a local minimum if $f(c)\le f(x)$ for all $x$ in a small interval around $c$. At $x=-1$, the function is defined by the first branch. Substituting $x=-1$ into $f(x)=k-2x$, we find the altitude at the junction:

For $f(-1)$ to be a local minimum, it must be the lowest point in the neighborhood. We already know that for $x<-1$, the function is decreasing towards $k+2$, so all values to the left are greater than $k+2$.

The real challenge is the right side. We need $f(-1)$ to be less than or equal to the values on the right. Mathematically, this means:

Substituting our values, we get the inequality $k+2\le 1$. Solving this is straightforward: subtract $2$ from both sides, and we arrive at:

We are given options: $0$, $-\frac{1}{2}$, $-1$, and $1$. Our condition $k\le -1$ tells us that any value of $k$ less than or equal to $-1$ will create a local minimum.

Among the choices, only $k=-1$ fits. When $k=-1$, the function becomes continuous, and the two lines meet perfectly at the point $(-1,1)$, forming a beautiful V-shaped valley.

You have successfully navigated the junction and found the floor of the function. Keep this logic in your toolkit—whenever you face a piecewise function, visualize the geometry before you touch the algebra.