Let's break this down step by step.

Given function:

You provided the function $f(x) = \int_1^x \frac{dt}{1+t^4}$, and we are tasked with showing that it is one-to-one, and then finding $(f^{-1})'(0)$.

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Step 1: Show that $f(x)$ is one-to-one

To prove that $f(x)$ is one-to-one, we need to show that it is either strictly increasing or strictly decreasing over its entire domain. The simplest way to do this is by checking the derivative of $f(x)$.

Find the derivative of $f(x)$:

Using the Fundamental Theorem of Calculus, we differentiate $f(x) = \int_1^x \frac{dt}{1+t^4}$:

$f'(x) = \frac{d}{dx} \left( \int_1^x \frac{dt}{1+t^4} \right)$ According to the Fundamental Theorem of Calculus:

$f'(x) = \frac{1}{1+x^4}$

Is $f'(x)$ positive or negative?

Note that $\frac{1}{1+x^4} > 0$ for all real $x$, since $1 + x^4 \geq 1$ for all $x$, and the denominator is always positive.

Thus, $f'(x) > 0$ for all $x \in (-\infty, \infty)$, meaning that $f(x)$ is strictly increasing on $(-\infty, \infty)$.

Since the derivative is positive for all $x$, this means that $f(x)$ is one-to-one, as it is strictly increasing and thus injective.

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Step 2: Find $(f^{-1})'(0)$

To find $(f^{-1})'(0)$, we can use the formula for the derivative of the inverse function:

$(f^{-1})'(y) = \frac{1}{f'(x)} \quad \text{where} \quad y = f(x)$

We are looking for $(f^{-1})'(0)$, so we need to determine $x$ such that $f(x) = 0$.

Find $x$ such that $f(x) = 0$:

From the definition of $f(x)$:

$f(x) = \int_1^x \frac{dt}{1+t^4}$

We want $f(x) = 0$, so we set the integral equal to zero:

$\int_1^x \frac{dt}{1+t^4} = 0$

Since the integrand $\frac{1}{1+t^4}$ is always positive, this integral will be zero only when the limits of integration are the same. Therefore, $f(x) = 0$ when $x = 1$.

Thus, $f(1) = 0$, so $f^{-1}(0) = 1$.

Now, compute $(f^{-1})'(0)$:

From the formula for the derivative of the inverse:

$(f^{-1})'(0) = \frac{1}{f'(1)}$

We already know that $f'(x) = \frac{1}{1+x^4}$. Therefore:

$f'(1) = \frac{1}{1+1^4} = \frac{1}{2}$

Thus:

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

Final Answer:

$(f^{-1})'(0) = 2$

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Would you like further details on any of these steps? Here are a few follow-up questions that may interest you:

1. How does the Fundamental Theorem of Calculus apply to integrals of functions like this?
2. How can we find the inverse of a function in general?
3. What would happen if $f(x)$ were not strictly increasing?
4. Why is $f(x)$ one-to-one on $(-\infty, \infty)$ in this case?
5. How do you compute the derivative of an inverse function in general?

Tip: When you're dealing with integrals and derivatives, remember that the Fundamental Theorem of Calculus provides a straightforward way to differentiate integrals where the upper limit of integration is a variable.