Let fx=(5x+1) then (x)is,
Let fx=(5x+1)2 ∀x>0, then f-1(x)is,
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Final inverse function: f⁻¹(x) = (√x - 1)/5, domain x > 1.
Start with the function: y = (5x + 1)², with domain x > 0.
Swap x and y to find the inverse: x = (5y + 1)².
Take the square root. Choose the positive root: √x = 5y + 1, because after swapping x represents the original function's output, which is > 1, so the right-hand side is positive.
Solve for y: y = (√x - 1)/5.
Note: The inverse's domain is x > 1 because the original function's range is (1, ∞).