In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:

b. Solve the equation y=f(x) for x as a function of y, and name the resulting inverse function g.

70. y= x³/(x²+1), -1 ≤ x ≤ 1, x_0=1/2

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Start with the given function: \(y = \frac{x^{3}}{x^{2} + 1}\), where \(-1 \leq x \leq 1\).

To find the inverse function \(g(y)\), we need to solve the equation \(y = \frac{x^{3}}{x^{2} + 1}\) for \(x\) in terms of \(y\).

Multiply both sides of the equation by \((x^{2} + 1)\) to clear the denominator: \(y(x^{2} + 1) = x^{3}\).

Rewrite this as a polynomial equation in \(x\): \(x^{3} - y x^{2} - y = 0\).

Solve this cubic equation for \(x\) as a function of \(y\). The solution(s) will define the inverse function(s) \(g(y)\). Since the original function is restricted to \(-1 \leq x \leq 1\), select the root that lies within this domain to define the inverse.

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Inverse Functions

An inverse function reverses the effect of the original function, swapping inputs and outputs. For a function f(x), its inverse g(y) satisfies g(f(x)) = x. Finding the inverse involves solving y = f(x) for x in terms of y, which may require algebraic manipulation and domain restrictions to ensure the inverse is well-defined.

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When the inverse function is not given explicitly, implicit differentiation helps find its derivative. The derivative of the inverse at a point is the reciprocal of the derivative of the original function at the corresponding point, i.e., g'(y) = 1 / f'(x). This relationship is crucial for understanding tangent lines and rates of change of inverse functions.

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Tangent line approximation uses the derivative at a point to approximate the function near that point with a linear function. For a function f at x₀, the tangent line is y = f(x₀) + f'(x₀)(x - x₀). This concept extends to inverse functions, allowing approximation of g(y) near y₀ using the derivative of g.

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