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Understanding whether a function is one-to-one is crucial for determining if it has an inverse. A function is considered one-to-one if every element in its range corresponds to exactly one element in its domain. In simpler terms, no two different input values (say, \(x_1\) and \(x_2\)) can yield the same output when plugged into the function. For the function \(f(x) = \sqrt[3]{5x+2}\), this property holds true because it is a cube root function. Cube root functions are strictly increasing. This means that as the input \(x\) increases, the output \(f(x)\) also increases without ever decreasing or plateauing. Since there is a continuous and one-to-one relationship between \(x\) and \(y\), it ensures that \(f(x)\) is one-to-one, and therefore, an inverse function exists.

Knowing the domain of an inverse function is important because it tells us the set of all permissible inputs for this inverse. The domain of \(f^{-1}(x)\) is actually the range of the original function \(f(x)\).For our function \(f(x) = \sqrt[3]{5x+2}\), the range is all real numbers because it's a cube root function, which covers every possible real number as you go along the x-axis. This occurs because cube root functions can handle negative, zero, and positive inputs, returning respective outputs across the entire function.Thus, the domain of the inverse function \(f^{-1}(x)\) is also all real numbers. This means \(f^{-1}(x) = \frac{x^3 - 2}{5}\) can accept any real number as an input.

To find the derivative of an inverse function, we use a special formula from calculus. When you have a one-to-one differentiable function like \(f(x) = \sqrt[3]{5x+2}\), its inverse \(f^{-1}(x)\) derivative at any point \(x\) is given by:\[(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}\]First, find the derivative of the original function \(f(x)\). For \(f(x) = (5x+2)^{1/3}\), the derivative is:\[ f'(x) = \frac{5}{3}(5x+2)^{-2/3} \]Next, substitute \(f^{-1}(x)\) in place of \(x\) in this derivative. Since \(f^{-1}(x) = \frac{x^3 - 2}{5}\), you plug this into \(f'(x)\) to get:\[ f'(f^{-1}(x)) = \frac{5}{3}x^{-2} \]Thus, the derivative of the inverse function is:\[ (f^{-1})'(x) = \frac{3x^2}{5} \]This derivative tells us how rapidly the inverse function \(f^{-1}(x)\) changes at any point \(x\). It is a crucial concept for evaluating the behavior and properties of inverse functions in calculus.