Algebra 2 Composition and Inverse Functions Quiz: 10 Questions with Explanations

A 7B 8C 6D 4

g(3) = 6, then f(6) = 7. Applying f first gives g(f(3)) = 8.

A fB Both at the same timeC NeitherD g

In f(g(x)), the inner function g is applied to x first.

A \(f^{-1}(x) = x + 9\)B \(f^{-1}(x) = x – 9\)C \(f^{-1}(x) = \frac{1}{x + 9}\)D \(f^{-1}(x) = 9 – x\)

To undo adding 9, subtract 9.

A \(m = 3c – 4\)B \(m = \frac{c – 4}{3}\)C \(m = \frac{c}{3} – 4\)D \(m = \frac{c + 4}{3}\)

Solve for m: subtract 4, then divide by 3. This is the inverse of the fare function. Dividing before subtracting gives c/3 – 4, which is wrong.

A \((3, 2)\)B \((2, -3)\)C \((-2, 3)\)D \((3, 3)\)

An inverse swaps the coordinates, so (2, 3) becomes (3, 2).

A \(f(x) = x^{2}\)B \(f(x) = |x|\)C \(f(x) = x^{4}\)D \(f(x) = x^{3}\)

Only x³ is one-to-one. The others give the same output for x and -x.

A Check that f(x) = g(x)B Check that f(g(x)) = x and g(f(x)) = xC Check that f(x) + g(x) = 0D Check that f(x) · g(x) = 1

Inverse functions undo each other, so both compositions give x.

A 2B 7C 3D 4

f(2) = 7, so the inverse sends 7 back to 2.

A \(f^{-1}(x) = \sqrt{x} – 2\)B \(f^{-1}(x) = \sqrt{x + 2}\)C \(f^{-1}(x) = \sqrt{x – 2}\)D \(f^{-1}(x) = x^{2} – 2\)

x = y² + 2 gives y² = x – 2, and y = √(x – 2) for the restricted domain.

A It is not continuousB Its graph does not pass through the originC A horizontal line meets it twice, so it is not one-to-oneD Its domain is restricted

Two inputs such as 2 and -2 give the same output, so the function fails the horizontal line test.