\[ A(x)=\int_{0}^{x} \frac{1}{1+t^{2}} d t \] for all \( x \). Note that we know (or willknow) from class that \( A(x)=\arctan (x) \). not to use this fact since you are tryipg to give alternate definitions for the trigon functions. Using this definition compute \( A(0) \) and \( A^{\prime}(x) \) for every \( x \). What is You may assume that \( p=\lim _{x \rightarrow \infty} A(x) \) exists. With this assumption comput lim \( m_{x \rightarrow-\infty} A(x) \) in terms of \( p \). Find the exact value of \( p \) without proof. Show that \( A(x) \) has an inverse function. Let \( T(x) \) be the inverse function State what the domain and range are for both \( A(x) \) and \( T(x) \). Compute the derivative \( T^{\prime}(x) \). (Your answer should be in terms of \( T(x) \).

Let's address each part of your problem step by step. ### 1. Compute \( A(0) \) and \( A^{\prime}(x) \) for Every \( x \) **a. Evaluating \( A(0) \):** \[ A(x) = \int_{0}^{x} \frac{1}{1+t^{2}} \, dt \] When \( x = 0 \): \[ A(0) = \int_{0}^{0} \frac{1}{1+t^{2}} \, dt = 0 \] **b. Finding \( A^{\prime}(x) \):** Using the **Fundamental Theorem of Calculus**, if \( F(x) = \int_{a}^{x} f(t) \, dt \), then: \[ F^{\prime}(x) = f(x) \] Applying this to \( A(x) \): \[ A^{\prime}(x) = \frac{1}{1+x^{2}} \] ### 2. Compute \( \lim_{x \to -\infty} A(x) \) in Terms of \( p \) Assume: \[ p = \lim_{x \to \infty} A(x) = \int_{0}^{\infty} \frac{1}{1+t^{2}} \, dt \] To find \( \lim_{x \to -\infty} A(x) \): \[ \lim_{x \to -\infty} A(x) = \lim_{x \to -\infty} \int_{0}^{x} \frac{1}{1+t^{2}} \, dt = -\int_{-\infty}^{0} \frac{1}{1+t^{2}} \, dt = -p \] ### 3. Find the Exact Value of \( p \) Without Proof The integral \( p = \int_{0}^{\infty} \frac{1}{1+t^{2}} \, dt \) is a well-known improper integral: \[ p = \frac{\pi}{2} \] ### 4. Show That \( A(x) \) Has an Inverse Function **Monotonicity and Continuity:** - **Continuity:** \( A(x) \) is continuous for all \( x \in \mathbb{R} \) because the integrand \( \frac{1}{1+t^{2}} \) is continuous. - **Monotonicity:** \( A^{\prime}(x) = \frac{1}{1+x^{2}} > 0 \) for all \( x \), implying \( A(x) \) is **strictly increasing**. Since \( A(x) \) is continuous and strictly increasing on \( \mathbb{R} \), it is a **bijective** function, and therefore, it **has an inverse function**. ### 5. Domains and Ranges of \( A(x) \) and Its Inverse \( T(x) \) Let \( T(x) \) be the inverse function of \( A(x) \). - **For \( A(x) \):** - **Domain:** \( \mathbb{R} \) (all real numbers) - **Range:** \( (-p, p) \) where \( p = \frac{\pi}{2} \) - **For \( T(x) \):** - **Domain:** \( (-p, p) \) - **Range:** \( \mathbb{R} \) (all real numbers) ### 6. Compute the Derivative \( T^{\prime}(x) \) in Terms of \( T(x) \) Using the property of inverse functions, if \( y = A(x) \) and \( x = T(y) \), then: \[ \frac{dy}{dx} = A^{\prime}(x) = \frac{1}{1+x^{2}} \] Taking the reciprocal to find \( \frac{dx}{dy} = T^{\prime}(y) \): \[ T^{\prime}(y) = 1 + x^{2} \] But since \( x = T(y) \), we substitute: \[ T^{\prime}(y) = 1 + [T(y)]^{2} \] **Final Answer:** - \( A(0) = 0 \). - For all real numbers \( x \), the derivative \( A^{\prime}(x) = \dfrac{1}{1 + x^{2}} \). - Assuming \( p = \lim_{x \to \infty} A(x) \), we have \( \lim_{x \to -\infty} A(x) = -p \). - The exact value of \( p \) is \( \dfrac{\pi}{2} \). - The function \( A(x) \) is invertible. - The domain of \( A(x) \) is all real numbers \( \mathbb{R} \), and its range is \( (-\pi/2,\, \pi/2) \). Conversely, the inverse function \( T(x) \) has domain \( (-\pi/2,\, \pi/2) \) and range \( \mathbb{R} \). - The derivative of the inverse function is \( T^{\prime}(x) = 1 + [T(x)]^{2} \).