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JcAlmighty

The answer is   \(f^{-1}(x) = \sqrt[3]{ \frac{x-3}{6}}\)

Our function is f(x) = 6x³ + 3. We need to find f⁻¹(x)
Step 1: Substitute y for f(x):
            y = 6x³ + 3
Step 2: Switch x and y because every (x, y) has its partner (y, x):
            x = 6y³ + 3 
Step 3: Solve for y:
                      \(6y^{3} =x-3 \\ y^{3}= \frac{x-3}{6} \\ y = \sqrt[3]{ \frac{x-3}{6}}\)
        \(f^{-1}(x) = \sqrt[3]{ \frac{x-3}{6}}\)

Answer: f-1(x) = Cube root of quantity x minus three divided by six.

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