If the function `f(x) = x^(3)-24x^(2)+6kx-8` is maximum at x = 2 then find value of k.

Answer

To solve the problem, we need to find the value of \( k \) such that the function \( f(x) = x^3 - 24x^2 + 6kx - 8 \) has a maximum at \( x = 2 \).

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

The first derivative of the function \( f(x) \) is given by:

\[ f'(x) = \frac{d}{dx}(x^3 - 24x^2 + 6kx - 8) \]

Calculating the derivative term by term:

- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( -24x^2 \) is \( -48x \).
- The derivative of \( 6kx \) is \( 6k \).
- The derivative of \( -8 \) is \( 0 \).

Thus, we have:

\[ f'(x) = 3x^2 - 48x + 6k \]

### Step 2: Set \( f'(2) = 0 \)

Since \( f(x) \) has a maximum at \( x = 2 \), we set the first derivative equal to zero at this point:

\[ f'(2) = 3(2)^2 - 48(2) + 6k = 0 \]

Calculating \( f'(2) \):

\[ f'(2) = 3(4) - 96 + 6k = 0 \]
\[ 12 - 96 + 6k = 0 \]
\[ 6k - 84 = 0 \]
\[ 6k = 84 \]
\[ k = \frac{84}{6} = 14 \]

### Step 3: Verify the second derivative condition \( f''(x) \)

Next, we need to ensure that \( f''(2) < 0 \) to confirm that there is indeed a maximum at \( x = 2 \).

First, we find the second derivative \( f''(x) \):

\[ f''(x) = \frac{d}{dx}(3x^2 - 48x + 6k) = 6x - 48 \]

Now, we evaluate \( f''(2) \):

\[ f''(2) = 6(2) - 48 = 12 - 48 = -36 \]

Since \( f''(2) < 0 \), this confirms that \( f(x) \) has a maximum at \( x = 2 \).

### Conclusion

Thus, the value of \( k \) is:

\[ \boxed{14} \]