If for the function 'f', given by :
`f(x)=kx^(2)+7x+4, f^(')(5)=97`, find 'k'.

Answer

To find the value of \( k \) for the function \( f(x) = kx^2 + 7x + 4 \) given that \( f'(5) = 97 \), we will follow these steps:

### Step 1: Find the derivative of the function
The function is given as:
\[ f(x) = kx^2 + 7x + 4 \]
To find the derivative \( f'(x) \), we differentiate each term:
\[ f'(x) = \frac{d}{dx}(kx^2) + \frac{d}{dx}(7x) + \frac{d}{dx}(4) \]
Using the power rule:
\[ f'(x) = 2kx + 7 + 0 = 2kx + 7 \]

### Step 2: Substitute \( x = 5 \) into the derivative
We know that \( f'(5) = 97 \). So we substitute \( x = 5 \) into the derivative:
\[ f'(5) = 2k(5) + 7 \]
This simplifies to:
\[ f'(5) = 10k + 7 \]

### Step 3: Set up the equation using the given information
Since we know \( f'(5) = 97 \), we can set up the equation:
\[ 10k + 7 = 97 \]

### Step 4: Solve for \( k \)
Now we will solve for \( k \):
\[ 10k = 97 - 7 \]
\[ 10k = 90 \]
\[ k = \frac{90}{10} = 9 \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{9} \]