Let `f : R to R ` be defined by
if `f(x)= {underset((2x + 3)" ", " "if" "x gt -1) (k-2x" "," "if" "x le -1).`
if f has a local minimum at x=-1 then a possible value of k is

Options

A. 0
B. `-(1)/(2)`
C. `-1`
D. 1

Answer

To solve the problem, we need to determine the value of \( k \) such that the function \( f(x) \) has a local minimum at \( x = -1 \). The function is defined as follows:

\[ f(x) = \begin{cases} 2x + 3 & \text{if } x > -1 \\ k - 2x & \text{if } x \leq -1 \end{cases} \]

### Step 1: Ensure Continuity at \( x = -1 \)

For \( f(x) \) to have a local minimum at \( x = -1 \), it must be continuous at that point. This means that the left-hand limit and the right-hand limit at \( x = -1 \) must be equal to the value of the function at \( x = -1 \).

**Right-hand limit** as \( x \) approaches \(-1\):
\[ \lim_{x \to -1^+} f(x) = 2(-1) + 3 = 1 \]

**Left-hand limit** as \( x \) approaches \(-1\):
\[ \lim_{x \to -1^-} f(x) = k - 2(-1) = k + 2 \]

Setting these two limits equal for continuity:
\[ k + 2 = 1 \]

### Step 2: Solve for \( k \)

Now, we solve the equation:
\[ k + 2 = 1 \]
Subtracting 2 from both sides:
\[ k = 1 - 2 \]
\[ k = -1 \]

### Conclusion

Thus, the possible value of \( k \) for which \( f \) has a local minimum at \( x = -1 \) is:
\[ \boxed{-1} \]