Let `f(x)=(g(x))/x w h e nx!=0` and `f(0)=0.` If `g(0)=g^(prime)(0)=0a n dg^(0)=17` then `f'(0)=` `3//4` b. `-1//2` c. `17//3` d. `17//2`

Options

A. `3//4`
B. `-1//2`
C. `17//3`
D. `17//2`

Explanation

To find \( f'(0) \) for the function \( f(x) = \frac{g(x)}{x} \) when \( x \neq 0 \) and \( f(0) = 0 \), we will use the given information and apply L'Hôpital's Rule due to the \( \frac{0}{0} \) form.

### Step 1: Write down the limit definition of the derivative at \( x = 0 \)

We start with the definition of the derivative:
\[ f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} \]
Since \( f(0) = 0 \), this simplifies to:
\[ f'(0) = \lim_{x \to 0} \frac{f(x)}{x} \]

### Step 2: Substitute \( f(x) \)

Substituting \( f(x) = \frac{g(x)}{x} \):
\[ f'(0) = \lim_{x \to 0} \frac{\frac{g(x)}{x}}{x} = \lim_{x \to 0} \frac{g(x)}{x^2} \]

### Step 3: Analyze the limit

As \( x \to 0 \), both \( g(0) = 0 \) and \( x^2 \to 0 \), resulting in a \( \frac{0}{0} \) form. We can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule

Using L'Hôpital's Rule, we differentiate the numerator and denominator:
\[ f'(0) = \lim_{x \to 0} \frac{g'(x)}{2x} \]
Again, as \( x \to 0 \), we have \( g'(0) = 0 \) and \( 2x \to 0 \), resulting in another \( \frac{0}{0} \) form. We apply L'Hôpital's Rule again.

### Step 5: Apply L'Hôpital's Rule again

Differentiating again gives:
\[ f'(0) = \lim_{x \to 0} \frac{g''(x)}{2} \]
Now, substituting \( g''(0) = 17 \):
\[ f'(0) = \frac{17}{2} \]

### Conclusion

Thus, the value of \( f'(0) \) is:
\[ \boxed{\frac{17}{2}} \]