Let `f(x) ={underset(ax, " "x lt0)(x^(2) , x ge0)`. Find real values of 'a' such that f(x) is strictly monotonically increasing at x=0

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} ax & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases} \]

We need to find the real values of \( a \) such that \( f(x) \) is strictly monotonically increasing at \( x = 0 \).

### Step 1: Find the derivative of \( f(x) \)

To determine if the function is monotonically increasing, we need to find the derivative \( f'(x) \).

- For \( x < 0 \):
\[ f'(x) = a \]

- For \( x \geq 0 \):
\[ f'(x) = 2x \]

### Step 2: Evaluate the derivative at \( x = 0 \)

Now, we need to check the behavior of \( f'(x) \) at \( x = 0 \):

- From the left (as \( x \) approaches 0 from the negative side):
\[ f'(0^-) = a \]

- From the right (as \( x \) approaches 0 from the positive side):
\[ f'(0^+) = 2 \cdot 0 = 0 \]

### Step 3: Ensure the function is strictly increasing at \( x = 0 \)

For \( f(x) \) to be strictly monotonically increasing at \( x = 0 \), the following condition must hold:
\[ f'(0^-) \geq 0 \quad \text{and} \quad f'(0^+) > 0 \]

From the above evaluations:
- \( f'(0^-) = a \)
- \( f'(0^+) = 0 \)

Since \( f'(0^+) = 0 \), we need \( f'(0^-) = a \) to be greater than or equal to 0:
\[ a \geq 0 \]

### Step 4: Ensure that the function does not decrease

For \( f(x) \) to be strictly increasing, we also need to ensure that \( a > 0 \) because if \( a = 0 \), the function would not be strictly increasing at \( x = 0 \).

### Conclusion

Thus, the real values of \( a \) such that \( f(x) \) is strictly monotonically increasing at \( x = 0 \) is:
\[ a > 0 \]