If (x) is differentiable in `[a,b]` such that `f(a)=2,f(b)=6,` then there exists at least one `c,a ltcleb,` such that `(b^(3)-a^(3))f'(c)=`

Options

A. `c^(2)`
B. `2c^(2)`
C. `-3c^(2)`
D. `12c^(2)`

Answer

To solve the problem, we will use Rolle's Theorem, which states that if a function is continuous on a closed interval and differentiable on the open interval, and if the function takes the same value at the endpoints of the interval, then there exists at least one point in the interval where the derivative of the function is zero.

### Step-by-Step Solution:

1. **Define the function**: We start by defining a new function \( h(x) \) as follows:
\[ h(x) = f(x) - f(a) + a \cdot (x^3 - a^3) \]
This function is constructed to help us apply Rolle's Theorem.

2. **Evaluate \( h(b) \)**: We calculate \( h(b) \):
\[ h(b) = f(b) - f(a) + a \cdot (b^3 - a^3) \]
Given \( f(a) = 2 \) and \( f(b) = 6 \), we substitute these values:
\[ h(b) = 6 - 2 + a \cdot (b^3 - a^3) = 4 + a \cdot (b^3 - a^3) \]

3. **Set \( h(b) = 0 \)**: For Rolle's Theorem to apply, we need \( h(a) = h(b) \). We set \( h(b) \) to zero:
\[ 4 + a \cdot (b^3 - a^3) = 0 \]
This implies:
\[ a \cdot (b^3 - a^3) = -4 \]
Therefore, we can solve for \( a \):
\[ a = \frac{-4}{b^3 - a^3} \]

4. **Evaluate \( h(a) \)**: Now we calculate \( h(a) \):
\[ h(a) = f(a) - f(a) + a \cdot (a^3 - a^3) = 0 \]
Thus, \( h(a) = 0 \).

5. **Apply Rolle's Theorem**: Since \( h(a) = h(b) = 0 \) and \( h(x) \) is continuous on \([a, b]\) and differentiable on \((a, b)\), by Rolle's Theorem, there exists at least one \( c \in (a, b) \) such that:
\[ h'(c) = 0 \]

6. **Differentiate \( h(x) \)**: Now we differentiate \( h(x) \):
\[ h'(x) = f'(x) + a \cdot 3x^2 \]
Setting \( h'(c) = 0 \):
\[ f'(c) + a \cdot 3c^2 = 0 \]
Rearranging gives:
\[ f'(c) = -a \cdot 3c^2 \]

7. **Multiply by \( (b^3 - a^3) \)**: Now we multiply both sides by \( (b^3 - a^3) \):
\[ (b^3 - a^3) f'(c) = -a \cdot 3c^2 (b^3 - a^3) \]
Thus, we have:
\[ (b^3 - a^3) f'(c) = 12c^2 \]

### Conclusion:
The final expression we derived is:
\[ (b^3 - a^3) f'(c) = 12c^2 \]
This shows that there exists at least one \( c \) in the interval \( (a, b) \) such that the equation holds.