Question

In Exercises \(61-66,\) find \(\left(f^{-1}\right)(a)\) for the function \(f\) and real number \(a\). $$ \begin{array}{ll} \text { Function } && \text { Real Number } \\ \hline f(x)=x^{3}+2 x-1 && a=2 \end{array} $$

Short answer

We can not solve \(\sqrt[3]{2 - 2y + 1}\) algebraically for \(y\). A numerical method such as Newton's Method or a suitable calculator or computer code can be used to find an approximate solution.

Step-by-step solution

Find the Inverse Function

The given function is \(f(x) = x^3 + 2x - 1\). To determine the inverse of this function, we will swap \(y\) for \(x\) and vice versa, and then solve for \(y\). This gives us \(x = y^3 + 2y - 1\). Solving for \(y\) gives us \(y = \sqrt[3]{x - 2y + 1}\). Because of the cubic function, this is a bit complex to solve for \(y\), and a standard algebraic method might not work. Here, we need to use numerical methods to find the inverse of the function. We note that \(f^{-1}(x) = \sqrt[3]{x - 2y + 1}\).

Substitute \(a\) into the Inverse Function

Next, we'll substitute \(a = 2\) into \(f^{-1}(x)\) to obtain \(f^{-1}(a)\). Plugging into our inverse function, we get: \(f^{-1}(2) = \sqrt[3]{2 - 2y + 1}\). This equation is complex to solve algebraically due to the cubic root. Normally, you'd isolate \(y\), but in this case, it's still contained within the cubic root on the right side of the equation. Hence, again, we might need to resort to numerical methods to solve for \(y\).

Solve Numerically

To solve this equation numerically for \(y\), we might use an iterative numerical method like Newton's Method or use a calculator or computer code that can handle complex numbers. Please note that these methods will not provide an exact answer, but an approximation.