Find the values of 'a' for which the function : `f(x)=x^(2)-2ax+6` is increasing when `x gt 0`.

Answer

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

### Step 1: Find the derivative of the function
The first step is to compute the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^2 - 2ax + 6) \]
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 increasing, the derivative must be greater than zero:

\[ f'(x) > 0 \]
Substituting the expression for \( f'(x) \):

\[ 2x - 2a > 0 \]

### Step 3: Simplify the inequality
We can simplify the inequality by dividing all terms by 2:

\[ x - a > 0 \]

### Step 4: Solve for \( x \)
Rearranging the inequality gives:

\[ x > a \]

### Step 5: Consider the condition \( x > 0 \)
Since we are interested in the interval where \( x > 0 \), we need to ensure that \( x > a \) holds true for all \( x > 0 \).

### Step 6: Analyze the inequality
For \( x > a \) to be true for all \( x > 0 \), the smallest value of \( x \), which is 0, must also satisfy the inequality:

\[ 0 > a \]

### Step 7: Conclusion
Thus, the condition for the function to be increasing when \( x > 0 \) is:

\[ a < 0 \]

### Final Answer
The values of 'a' for which the function \( f(x) = x^2 - 2ax + 6 \) is increasing when \( x > 0 \) are:

\[ a < 0 \]