Find the value of a for which the function `f(x) =x^(2)-2ax+6,x gt 0` is strictly increasing.

Answer

To find the value of \( a \) for which the function \( f(x) = x^2 - 2ax + 6 \) is strictly increasing for \( x > 0 \), we will follow these steps:

### Step 1: Find the derivative of the function
To determine when the function is strictly increasing, we first need to find its derivative \( f'(x) \).

\[ f'(x) = \frac{d}{dx}(x^2 - 2ax + 6) \]

Using the power rule and the constant multiple rule, we get:

\[ f'(x) = 2x - 2a \]

### Step 2: Set the derivative greater than zero
For the function to be strictly increasing, the derivative must be greater than zero:

\[ f'(x) > 0 \]

Substituting the expression we found for \( f'(x) \):

\[ 2x - 2a > 0 \]

### Step 3: Simplify the inequality
We can simplify the inequality:

\[ 2x > 2a \]

Dividing both sides by 2:

\[ x > a \]

### Step 4: Consider the domain of \( x \)
Since we are given that \( x > 0 \), we can conclude that for the function to be strictly increasing, \( a \) must be less than \( x \).

### Step 5: Find the maximum possible value of \( a \)
Since \( x \) can take any positive value greater than 0, the maximum value of \( a \) must be less than or equal to the smallest value of \( x \), which is approaching 0 (but not including 0). Therefore, we can conclude:

\[ a < 0 \]

### Conclusion
The value of \( a \) for which the function \( f(x) \) is strictly increasing for \( x > 0 \) is:

\[ a \leq 0 \]