Find the inverse of any function with step-by-step solutions. Supports linear, power, and rational functions.
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The inverse function f⁻¹(x) "undoes" f(x). If f(a) = b, then f⁻¹(b) = a. Graphically, it's the reflection of f(x) over the line y = x.
Replace f(x) with y, swap x and y, then solve for y. The result is f⁻¹(x). This works for one-to-one functions.
For f(x) = ax + b, the inverse is f⁻¹(x) = (x - b) / a. Linear functions (with a ≠ 0) always have inverses.
For f(x) = x², the inverse is f⁻¹(x) = ±√x. You must restrict the domain (x ≥ 0) to make it a true function.
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To find the inverse: (1) Write y = f(x), (2) Swap x and y, (3) Solve for y. The result is f⁻¹(x). For example, if f(x) = 2x + 3, swap to get x = 2y + 3, then solve: y = (x - 3)/2, so f⁻¹(x) = (x - 3)/2.
The inverse of f(x) = x² is f⁻¹(x) = ±√x. Since x² is not one-to-one over all reals, you restrict the domain to x ≥ 0, giving f⁻¹(x) = √x.
No. Only one-to-one (injective) functions have inverses. A function is one-to-one if each output corresponds to exactly one input. Use the horizontal line test: if any horizontal line crosses the graph more than once, the function is not one-to-one.
For f(x) = (ax + b)/(cx + d), the inverse is f⁻¹(x) = (dx - b)/(a - cx). Cross-multiply x(cx + d) = ax + b, collect y terms, and solve. The domain excludes x = a/c.
Verify by checking that f(f⁻¹(x)) = x and f⁻¹(f(x)) = x. If both compositions equal x, the inverse is correct.