104–107. Functions from derivatives Find the function f with the following properties.




h'(x) = (x⁴ -2) /(1 + x²) ; h (1) = -(2/3) 

Verified step by step guidance

1. Step 1: Recognize that the problem involves finding the original function h(x) given its derivative h'(x) = (x⁴ - 2) / (1 + x²) and an initial condition h(1) = -2/3. This is a problem of integration and using the constant of integration.
2. Step 2: Set up the integral to find h(x). To reverse the derivative, integrate h'(x): ∫((x⁴ - 2) / (1 + x²)) dx. This will give the general form of h(x) plus a constant of integration, C.
3. Step 3: Break the integral into manageable parts. Notice that the numerator (x⁴ - 2) can be split into two terms: x⁴ and -2. Rewrite the integral as ∫(x⁴ / (1 + x²)) dx - ∫(2 / (1 + x²)) dx.
4. Step 4: Solve each integral separately. For ∫(x⁴ / (1 + x²)) dx, use polynomial division or substitution techniques to simplify. For ∫(2 / (1 + x²)) dx, recognize it as a standard integral that results in 2 * arctan(x).
5. Step 5: Combine the results of the integrals and add the constant of integration, C. Use the initial condition h(1) = -2/3 to solve for C by substituting x = 1 into the expression for h(x).

Key Concepts

Derivatives

A derivative represents the rate of change of a function with respect to its variable. It is a fundamental concept in calculus that allows us to understand how a function behaves locally. The notation h'(x) indicates the derivative of the function h with respect to x, which can be used to find slopes of tangent lines and optimize functions.

Integration

Integration is the process of finding the antiderivative of a function, essentially reversing differentiation. It allows us to recover the original function from its derivative. In this context, to find the function h from its derivative h'(x), we will need to perform integration, which may involve applying techniques such as substitution or integration by parts.

Initial Conditions

Initial conditions are specific values that help determine a unique solution to a differential equation. In this problem, the condition h(1) = -2/3 provides a point through which the function h must pass. This information is crucial for solving the integral obtained from the derivative, as it allows us to find the constant of integration and fully define the function.