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Functions, domains, ranges and linear inverses — 0580

Exchange the permitted inputs and attained outputs when reversing a one-to-one rule.

Extended · Syllabus E2.13

Notes and key methods

A function gives one output for every allowed input. Its domain is the input set and its range consists of the outputs it actually attains. For a strictly increasing linear rule on a closed interval, the endpoint images give the included range endpoints.

To find an inverse, write \(\displaystyle y=f(x)\), solve for x, then interchange input and output. A one-to-one rule has exactly one original input for each attained output. The inverse domain is the original range; its range is the original domain.

Check the conditions

Three worked examples

Example 1 · Functions, domains, ranges and linear inverses

\(f(x)=3 x + 4\) with domain \(\left[1, 5\right]\).

  1. State the range of f and justify why it has an inverse.
  2. Find the inverse rule and state its domain and range.
  3. Find \(f^{-1}(10)\) and check it in f.
  4. Find the gradient of the inverse rule.
Show the reasoning

(a)

The positive gradient 3 makes f strictly increasing. Its endpoint outputs are 7 and 19, and both are attained.

Answer: \(\left[7, 19\right]\); f is one-to-one on its stated domain.

(b)

Solve \(y=3 x + 4\) for x, interchange the variables, and exchange the original domain and range.

Answer: \(\displaystyle f^{-1}(x)=\frac{x}{3} - \frac{4}{3}\); domain \(\left[7, 19\right]\), range \(\left[1, 5\right]\).

(c)

\(f(2)=10\) verifies the reverse calculation. The input 2 lies in the original domain.

Answer: \(2\)

(d)

Use the stated input interval, evaluate the linear rule or reverse its operations for an inverse input.

\(\displaystyle \frac{1}{3}\)

Answer: \(\displaystyle \frac{1}{3}\).

Example 2 · Functions, domains, ranges and linear inverses

\(f(x)=4 x + 5\) with domain \(\left[2, 7\right]\).

  1. State the range of f and justify why it has an inverse.
  2. Find the inverse rule and state its domain and range.
  3. Find \(f^{-1}(17)\) and check it in f.
  4. Find the gradient of the inverse rule.
Show the reasoning

(a)

The positive gradient 4 makes f strictly increasing. Its endpoint outputs are 13 and 33, and both are attained.

Answer: \(\left[13, 33\right]\); f is one-to-one on its stated domain.

(b)

Solve \(y=4 x + 5\) for x, interchange the variables, and exchange the original domain and range.

Answer: \(\displaystyle f^{-1}(x)=\frac{x}{4} - \frac{5}{4}\); domain \(\left[13, 33\right]\), range \(\left[2, 7\right]\).

(c)

\(f(3)=17\) verifies the reverse calculation. The input 3 lies in the original domain.

Answer: \(3\)

(d)

Use the stated input interval, evaluate the linear rule or reverse its operations for an inverse input.

\(\displaystyle \frac{1}{4}\)

Answer: \(\displaystyle \frac{1}{4}\).

Example 3 · Functions, domains, ranges and linear inverses

\(f(x)=5 x + 6\) with domain \(\left[-2, 4\right]\).

  1. State the range of f and justify why it has an inverse.
  2. Find the inverse rule and state its domain and range.
  3. Find \(f^{-1}(1)\) and check it in f.
  4. Find the gradient of the inverse rule.
Show the reasoning

(a)

The positive gradient 5 makes f strictly increasing. Its endpoint outputs are -4 and 26, and both are attained.

Answer: \(\left[-4, 26\right]\); f is one-to-one on its stated domain.

(b)

Solve \(y=5 x + 6\) for x, interchange the variables, and exchange the original domain and range.

Answer: \(\displaystyle f^{-1}(x)=\frac{x}{5} - \frac{6}{5}\); domain \(\left[-4, 26\right]\), range \(\left[-2, 4\right]\).

(c)

\(f(-1)=1\) verifies the reverse calculation. The input -1 lies in the original domain.

Answer: \(-1\)

(d)

Use the stated input interval, evaluate the linear rule or reverse its operations for an inverse input.

\(\displaystyle \frac{1}{5}\)

Answer: \(\displaystyle \frac{1}{5}\).

Common mistakes

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