Let `f` be a real-valued function defined on the inverval `(-1,1)` such that `e^(-x)f(x)=2+int_0^xsqrt(t^4+1)dt ,` for all, `x in (-1,1)a n dl e tf^(-1)` be the inverse function of `fdot` Then `(f^(-1))^'(2)` is equal to 1 (b) `1/3` (c) `1/2` (d) `1/e`

Options

A. `1`
B. `1//3`
C. `1//2`
D. `1//e`

Explanation

To solve the problem, we need to find the value of \( (f^{-1})'(2) \) given the function \( f \) defined by the equation:

\[ e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \]

### Step 1: Find \( f(0) \)

First, we will evaluate \( f(0) \) by substituting \( x = 0 \) into the given equation:

\[ e^{0} f(0) = 2 + \int_0^0 \sqrt{t^4 + 1} \, dt \]

The integral from 0 to 0 is 0, so we have:

\[ f(0) = 2 \]

### Step 2: Find \( f^{-1}(2) \)

Since \( f(0) = 2 \), we can conclude that:

\[ f^{-1}(2) = 0 \]

### Step 3: Differentiate the given equation

Next, we differentiate both sides of the equation \( e^{-x} f(x) = 2 + \int_0^x \sqrt{t^4 + 1} \, dt \) with respect to \( x \):

Using the product rule on the left side:

\[ \frac{d}{dx}(e^{-x} f(x)) = -e^{-x} f(x) + e^{-x} f'(x) \]

For the right side, we apply the Fundamental Theorem of Calculus:

\[ \frac{d}{dx}\left(2 + \int_0^x \sqrt{t^4 + 1} \, dt\right) = \sqrt{x^4 + 1} \]

Setting both sides equal gives us:

\[ -e^{-x} f(x) + e^{-x} f'(x) = \sqrt{x^4 + 1} \]

### Step 4: Evaluate at \( x = 0 \)

Now, we substitute \( x = 0 \):

\[ -e^{0} f(0) + e^{0} f'(0) = \sqrt{0^4 + 1} \]

This simplifies to:

\[ -f(0) + f'(0) = 1 \]

Since \( f(0) = 2 \):

\[ -2 + f'(0) = 1 \]

Thus:

\[ f'(0) = 3 \]

### Step 5: Use the inverse function theorem

From the inverse function theorem, we know:

\[ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} \]

Specifically, we need to find \( (f^{-1})'(2) \):

\[ (f^{-1})'(2) = \frac{1}{f'(f^{-1}(2))} = \frac{1}{f'(0)} \]

Substituting the value we found for \( f'(0) \):

\[ (f^{-1})'(2) = \frac{1}{3} \]

### Final Answer

Thus, the value of \( (f^{-1})'(2) \) is:

\[ \boxed{\frac{1}{3}} \]