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Question: Let f be a real-valued function defined on the interval (-1, 1) such that $e^{-x}f(x) = 2 + \int_{0}...

Let f be a real-valued function defined on the interval (-1, 1) such that e−xf(x)=2+∫0xt4+1dte^{-x}f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} dt, for all x∈(−1,1)x \in (-1,1), and let f−1f^{-1} be the inverse function of f. Then (f−1)′(2)(f^{-1})'(2) is equal to:

A

1

B

13\frac{1}{3}

C

12\frac{1}{2}

D

1e\frac{1}{e}

Answer

1/3

Explanation

Solution

The problem asks us to find the derivative of the inverse function, (f−1)′(2)(f^{-1})'(2), given a functional equation for f(x)f(x).

1. Understand the relationship between f(x)f(x) and f−1(x)f^{-1}(x):

Let y=f(x)y = f(x). Then x=f−1(y)x = f^{-1}(y). The derivative of the inverse function is given by the formula: (f−1)′(y)=1f′(x)(f^{-1})'(y) = \frac{1}{f'(x)} where y=f(x)y = f(x).

In this problem, we need to find (f−1)′(2)(f^{-1})'(2). This means we need to find the value of xx such that f(x)=2f(x) = 2. Let this value be x0x_0. Then we will compute 1f′(x0)\frac{1}{f'(x_0)}.

2. Find xx such that f(x)=2f(x) = 2:

The given equation is: e−xf(x)=2+∫0xt4+1dte^{-x}f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} dt

Let's try a simple value for xx, specifically x=0x=0. Substitute x=0x=0 into the equation: e−0f(0)=2+∫00t4+1dte^{-0}f(0) = 2 + \int_{0}^{0} \sqrt{t^4 + 1} dt 1⋅f(0)=2+01 \cdot f(0) = 2 + 0 f(0)=2f(0) = 2

So, when f(x)=2f(x) = 2, we have x=0x = 0. This means x0=0x_0 = 0. Therefore, we need to calculate (f−1)′(2)=1f′(0)(f^{-1})'(2) = \frac{1}{f'(0)}.

3. Find f′(x)f'(x) by differentiating the given equation:

Differentiate both sides of the equation e−xf(x)=2+∫0xt4+1dte^{-x}f(x) = 2 + \int_{0}^{x} \sqrt{t^4 + 1} dt with respect to xx.

Left side: Use the product rule ddx(uv)=u′v+uv′\frac{d}{dx}(uv) = u'v + uv' where u=e−xu = e^{-x} and v=f(x)v = f(x). ddx(e−xf(x))=−e−xf(x)+e−xf′(x)\frac{d}{dx}(e^{-x}f(x)) = -e^{-x}f(x) + e^{-x}f'(x)

Right side: Use the Fundamental Theorem of Calculus, ddx∫axg(t)dt=g(x)\frac{d}{dx} \int_{a}^{x} g(t) dt = g(x). ddx(2+∫0xt4+1dt)=0+x4+1\frac{d}{dx} \left(2 + \int_{0}^{x} \sqrt{t^4 + 1} dt\right) = 0 + \sqrt{x^4 + 1}

Equating the derivatives of both sides: −e−xf(x)+e−xf′(x)=x4+1-e^{-x}f(x) + e^{-x}f'(x) = \sqrt{x^4 + 1}

4. Find f′(0)f'(0):

Substitute x=0x=0 into the differentiated equation: −e−0f(0)+e−0f′(0)=04+1-e^{-0}f(0) + e^{-0}f'(0) = \sqrt{0^4 + 1} −1⋅f(0)+1⋅f′(0)=1-1 \cdot f(0) + 1 \cdot f'(0) = \sqrt{1} −f(0)+f′(0)=1-f(0) + f'(0) = 1

We already found that f(0)=2f(0) = 2. Substitute this value: −2+f′(0)=1-2 + f'(0) = 1 f′(0)=1+2f'(0) = 1 + 2 f′(0)=3f'(0) = 3

5. Calculate (f−1)′(2)(f^{-1})'(2):

Using the formula (f−1)′(2)=1f′(0)(f^{-1})'(2) = \frac{1}{f'(0)}: (f−1)′(2)=13(f^{-1})'(2) = \frac{1}{3}