Let `f : R to R ` be a continuously differentiable function such that `f(2) = 6 and f'(2) = 1/48 *` If `int_(6)^(f(x)) 4t^(3) dt = (x-2) g(x)" than" lim_( x to 2) g(x) ` is equal to

Options

A. 18
B. 24
C. 12
D. 36

Explanation

To solve the problem, we need to find the limit of \( g(x) \) as \( x \) approaches 2, given the equation:

\[ \int_{6}^{f(x)} 4t^3 \, dt = (x-2) g(x) \]

### Step 1: Evaluate the integral

First, we need to evaluate the integral:

\[ \int_{6}^{f(x)} 4t^3 \, dt \]

Using the fundamental theorem of calculus, we can compute this integral:

\[ \int 4t^3 \, dt = t^4 + C \]

Thus,

\[ \int_{6}^{f(x)} 4t^3 \, dt = f(x)^4 - 6^4 \]

Calculating \( 6^4 \):

\[ 6^4 = 1296 \]

So we have:

\[ \int_{6}^{f(x)} 4t^3 \, dt = f(x)^4 - 1296 \]

### Step 2: Substitute into the equation

Now substituting this back into our equation gives:

\[ f(x)^4 - 1296 = (x-2) g(x) \]

### Step 3: Analyze the limit as \( x \to 2 \)

As \( x \to 2 \):

- We know \( f(2) = 6 \), so \( f(x) \to 6 \).
- Therefore, \( f(x)^4 \to 6^4 = 1296 \).

This means that both sides of the equation approach 0 as \( x \to 2 \):

\[ 0 = 0 \cdot g(2) \]

This is an indeterminate form \( 0 = 0 \), so we can apply L'Hôpital's Rule.

### Step 4: Apply L'Hôpital's Rule

We differentiate both sides with respect to \( x \):

\[ \frac{d}{dx}(f(x)^4 - 1296) = \frac{d}{dx}((x-2) g(x)) \]

Using the chain rule on the left side:

\[ 4f(x)^3 f'(x) \]

On the right side, we apply the product rule:

\[ g(x) + (x-2) g'(x) \]

Setting the derivatives equal gives:

\[ 4f(x)^3 f'(x) = g(x) + (x-2) g'(x) \]

### Step 5: Evaluate at \( x = 2 \)

Now we substitute \( x = 2 \):

- \( f(2) = 6 \)
- \( f'(2) = \frac{1}{48} \)

Thus:

\[ 4(6)^3 \left(\frac{1}{48}\right) = g(2) + 0 \cdot g'(2) \]

Calculating \( 4(6)^3 \):

\[ 6^3 = 216 \quad \Rightarrow \quad 4 \cdot 216 = 864 \]

So we have:

\[ 864 \cdot \frac{1}{48} = g(2) \]

Calculating \( \frac{864}{48} \):

\[ g(2) = 18 \]

### Final Answer

Thus, the limit is:

\[ \lim_{x \to 2} g(x) = 18 \]