Finding the Inverse of a Linear Function

Find the inverse of $f(x) = 2x + 3$

Replace f(x) with y: $y = 2x + 3$ = Standard form

Swap x and y: $x = 2y + 3$ = Variables swapped

Solve for y: subtract 3: $x - 3 = 2y$ = Isolate the y term

Divide by 2: $y = \frac{x - 3}{2}$ = y is isolated

Write as inverse function: $f^{-1}(x) = \frac{x - 3}{2}$ = Inverse found

Answer: $f^{-1}(x) = \frac{x - 3}{2}$

Verifying Two Functions are Inverses

Verify that $f(x) = 3x - 6$ and $g(x) = \frac{x + 6}{3}$ are inverses

Check f(g(x)): $f(g(x)) = f\left(\frac{x + 6}{3}\right) = 3 \cdot \frac{x + 6}{3} - 6$ = Substitute g(x) into f

Simplify f(g(x)): $= (x + 6) - 6 = x$ = $f(g(x)) = x$ \checkmark

Check g(f(x)): $g(f(x)) = g(3x - 6) = \frac{(3x - 6) + 6}{3}$ = Substitute f(x) into g

Simplify g(f(x)): $= \frac{3x}{3} = x$ = $g(f(x)) = x$ \checkmark

Conclusion: Both compositions equal $x$ = They are inverses

Answer: Yes, $f$ and $g$ are inverse functions because $f(g(x)) = g(f(x)) = x$

Finding the Inverse of a Rational Function

Find the inverse of $f(x) = \frac{2x + 1}{x - 3}$, where $x \neq 3$

Replace f(x) with y: $y = \frac{2x + 1}{x - 3}$ = Standard form

Swap x and y: $x = \frac{2y + 1}{y - 3}$ = Variables swapped

Multiply both sides by (y - 3): $x(y - 3) = 2y + 1$ = Clear the fraction

Expand: $xy - 3x = 2y + 1$ = Distribute x

Collect y terms on one side: $xy - 2y = 3x + 1$ = Move y terms left

Factor out y: $y(x - 2) = 3x + 1$ = Common factor

Solve for y: $y = \frac{3x + 1}{x - 2}$ = Divide by (x - 2)

Write as inverse: $f^{-1}(x) = \frac{3x + 1}{x - 2}$, where $x \neq 2$ = Include domain restriction

Answer: $f^{-1}(x) = \frac{3x + 1}{x - 2}$, where $x \neq 2$

Mistake: Confusing $f^{-1}(x)$ with $\frac{1}{f(x)}$

Why: The notation $f^{-1}$ means the inverse function, NOT the reciprocal. $f^{-1}(x)$ undoes $f(x)$, while $\frac{1}{f(x)}$ is just one divided by $f(x)$.

Correct: $f^{-1}(x)$ is the inverse function. For the reciprocal, write $\frac{1}{f(x)}$ or $(f(x))^{-1}$.

Mistake: Forgetting to swap x and y before solving

Why: The swap is essential because the inverse reverses inputs and outputs. Without swapping, you'll just solve for x in terms of y, not find the inverse.

Correct: Always follow the steps: write y = f(x), swap to get x = f(y), then solve for y.

Mistake: Assuming all functions have inverses

Why: Only one-to-one functions (passing the horizontal line test) have inverses. Functions like $f(x) = x^2$ fail because two inputs give the same output.

Correct: Check that the function is one-to-one first. For $f(x) = x^2$, restrict the domain to $x \geq 0$ to create an inverse.

Mistake: Forgetting domain restrictions for the inverse

Why: The domain of $f^{-1}$ is the range of $f$, and vice versa. If the original function excludes certain values, the inverse will have different restrictions.

Correct: Always state the domain of the inverse. If $f(x) = \frac{1}{x-2}$ has domain $x \neq 2$, then $f^{-1}$ will have a different restriction.

Temperature Conversion

Converting between Celsius and Fahrenheit uses inverse functions.

The formula $F = \frac{9}{5}C + 32$ converts Celsius to Fahrenheit. Its inverse $C = \frac{5}{9}(F - 32)$ converts back.

Decryption in Cryptography

Simple encryption functions need inverses for decryption. In a Caesar cipher, if encryption shifts letters by 3, decryption shifts by -3.

If encryption is $E(x) = x + 3 \pmod{26}$, decryption is $D(x) = x - 3 \pmod{26}$

Q: What is the difference between $f^{-1}(x)$ and $f(x)^{-1}$?

A: $f^{-1}(x)$ is the inverse function - it undoes $f$. $f(x)^{-1}$ or $(f(x))^{-1}$ is the reciprocal $\frac{1}{f(x)}$. They are completely different! For example, if $f(x) = 2x$, then $f^{-1}(x) = \frac{x}{2}$, but $f(x)^{-1} = \frac{1}{2x}$.

Q: How do I know if a function has an inverse?

A: A function has an inverse if and only if it is one-to-one (each output comes from exactly one input). Use the horizontal line test: if any horizontal line crosses the graph more than once, the function does NOT have an inverse.

Q: Why do we swap x and y when finding the inverse?

A: Because the inverse swaps inputs and outputs. If $f$ takes 2 and gives 7, then $f^{-1}$ takes 7 and gives 2. Swapping x and y in the equation captures this reversal of roles.