1. Given function

We want $f$ to have a local minimum at $x=-1$.

2. Value of the function at $x=-1$

Since the first branch includes $x=-1$,

3. Behavior on the left and right of $x=-1$

• For $x\le -1$, $f(x)=k-2x,$ which has slope $-2<0$. So as $x$ increases toward $-1$ from the left, $f(x)$ decreases. Hence for $x<-1$ close to $-1$, $f(x)>f(-1).$

• For $x>-1$, $f(x)=2x+3,$ which has slope $2>0$. So as $x$ increases to the right of $-1$, $f(x)$ increases. Therefore near $-1$ on the right, the smallest nearby value on that branch is the right-hand limit: $\lim_{x\to -1^+}f(x)=2(-1)+3=1.$

For $x=-1$ to be a local minimum, we need nearby values on both sides to be at least $f(-1)$. The left side already satisfies this automatically. So we only need

Thus,

4. Check the options

• A: $k=0$ $k+2=2>1$ Right-side nearby values are less than $f(-1)$, so not a local minimum.

• B: $k=-\frac12$ $k+2=\frac32>1$ Not a local minimum.

• C: $k=-1$ $k+2=1$ Then $f(-1)=1$, and nearby values on both sides are $\ge 1$. So this works.

• D: $k=1$ $k+2=3>1$ Not a local minimum.

5. Conclusion

A possible value of $k$ is

So the correct option is C.