1. Graph ( f(x) = x^2 ). 2. Apply the horizontal line test. 3. Since a horizontal line can intersect the graph more than once, ( f(x) = x^2 ) is not invertible unless the domain is restricted.

1. Replace ( f(x) ) with ( y ): ( y = 3x + 2 ). 2. Swap ( x ) and ( y ): ( x = 3y + 2 ). 3. Solve for ( y ): ( y = (x - 2) / 3 ). 4. Replace ( y ) with ( f^{-1}(x) ): ( f^{-1}(x) = (x - 2) / 3 ).

1. For ( f(x) ): Domain is ( x geq 1 ), Range is ( y geq 0 ). 2. Find the inverse: ( f^{-1}(x) = x^2 + 1 ). 3. For ( f^{-1}(x) ): Domain is ( x geq 0 ), Range is ( y geq 1 ).

1. Compute ( f(f^{-1}(x)) ): ( f((x - 1) / 2) = 2((x - 1) / 2) + 1 = x - 1 + 1 = x ). 2. Compute ( f^{-1}(f(x)) ): ( f^{-1}(2x + 1) = ((2x + 1) - 1) / 2 = 2x / 2 = x ). 3. Since both compositions equal ( x ), the functions are inverses.

1. Replace ( f(x) ) with ( y ): ( y = x^3 ). 2. Swap ( x ) and ( y ): ( x = y^3 ). 3. Solve for ( y ): ( y = sqrt[3]{x} ). 4. Replace ( y ) with ( f^{-1}(x) ): ( f^{-1}(x) = sqrt[3]{x} ).

1. Replace ( f(x) ) with ( y ): ( y = frac{1}{x} ). 2. Swap ( x ) and ( y ): ( x = frac{1}{y} ). 3. Solve for ( y ): ( y = frac{1}{x} ). 4. Replace ( y ) with ( f^{-1}(x) ): ( f^{-1}(x) = frac{1}{x} ).

1. Replace ( f(x) ) with ( y ): ( y = e^x ). 2. Swap ( x ) and ( y ): ( x = e^y ). 3. Solve for ( y ): ( y = ln(x) ). 4. Replace ( y ) with ( f^{-1}(x) ): ( f^{-1}(x) = ln(x) ).

1. Find the range of ( f(x) ): Range is ( y geq 0 ). 2. The domain of ( f^{-1}(x) ) is the range of ( f(x) ), so the domain of ( f^{-1}(x) ) is ( x geq 0 ).

1. Choose either ( x geq 0 ) or ( x leq 0 ). 2. If ( x geq 0 ), the function is one-to-one and invertible. 3. If ( x leq 0 ), the function is one-to-one and invertible.

1. Find ( f^{-1}(x) ): ( f^{-1}(x) = frac{x + 2}{5} ). 2. Substitute ( x = 7 ) into ( f^{-1}(x) ): ( f^{-1}(7) = frac{7 + 2}{5} = frac{9}{5} ).

Function: Maps x to y | Inverse: Maps y to x

Function: Domain is x-values, Range is y-values | Inverse: Domain is y-values, Range is x-values

One-to-one: Each y-value corresponds to one x-value, invertible | Not one-to-one: Some y-values correspond to multiple x-values, not invertible without domain restriction

Function: Original graph | Inverse: Reflection of the original graph across the line ( y = x )

Linear: Straightforward algebraic manipulation | Quadratic: Requires domain restriction to ensure one-to-one property, more complex algebra

( f(x) = x^2 ): Requires restriction to ( x geq 0 ) or ( x leq 0 ) | ( f(x) = x^3 ): No restriction needed, already one-to-one

Horizontal Line Test: Checks if a function is one-to-one (invertible) | Vertical Line Test: Checks if a relation is a function

( f(x) = 2x + 3 ): Simple algebraic steps | ( f(x) = sqrt{x} ): Squaring operation needed, domain consideration

Both are equal to x, but ( f(f^{-1}(x)) ) means applying the inverse first, then the function, while ( f^{-1}(f(x)) ) means applying the function first, then the inverse.

( f(x) = sin(x) ): Not invertible without domain restriction | ( f(x) = arcsin(x) ): Invertible by definition, restricted domain

A function where each output (y-value) corresponds to only one input (x-value).

A function that has an inverse function. It must be one-to-one and have an unrestricted domain.

The inverse of a function ( f(x) ) is written as ( f^{-1}(x) ).

The set of all possible input values (x-values) for which the function is defined.

The set of all possible output values (y-values) that the function can produce.

A test to determine if a function is one-to-one. If any horizontal line intersects the graph more than once, the function is not one-to-one.

The function's domain isn't limited in a way that prevents it from having an inverse; it covers all possible input values.

If ( f(a) = b ), then ( f^{-1}(b) = a ). The input and output pairs are flipped.

If a function isn't one-to-one, its inverse might not be a function (it could have multiple outputs for one input).

The graph of ( f^{-1}(x) ) is a reflection of ( f(x) ) across the line ( y = x ).