Inverse functions and one-to-one functions

Key takeaway: These exercises focus on inverse functions, composition, and identifying one-to-one behavior. The key ideas are to swap variables, solve for the new output, and check the domain when a function is not one-to-one on all real numbers.

6. Show that $f(x)=a-x$ is its own inverse

Let

$y=a-x.$

Swap $x$ and $y$:

$x=a-y.$

Solve for $y$:

$y=a-x.$

So the inverse is the same function:

$f^{-1}(x)=a-x.$

Therefore, $f$ is its own inverse for every real number $a$.

7–12. Finding inverses

Use the same procedure each time:

1. Write $y=f(x)$.
2. Swap $x$ and $y$.
3. Solve for $y$.

• 7. $f(x)=x+3$
  $f^{-1}(x)=x-3$
• 8. $f(x)=x+5$
  $f^{-1}(x)=x-5$
• 9. $f(x)=2-x$
  $f^{-1}(x)=2-x$
• 10. $f(x)=3-x$
  $f^{-1}(x)=3-x$
• 11. $f(x)=\frac{x}{x+2}$
  Start with $y=\frac{x}{x+2}$, then solve for $x$:

$y(x+2)=x \Rightarrow yx+2y=x \Rightarrow yx-x=-2y$

$x(y-1)=-2y \Rightarrow x=\frac{-2y}{y-1}=\frac{2y}{1-y}$

So

$f^{-1}(x)=\frac{2x}{1-x}.$

• 12. $f(x)=\frac{2x+3}{5x+4}$
  Set $y=\frac{2x+3}{5x+4}$ and solve:

$y(5x+4)=2x+3$

$5yx+4y=2x+3$

$5yx-2x=3-4y$

$x(5y-2)=3-4y$

$x=\frac{3-4y}{5y-2}$

Thus

$f^{-1}(x)=\frac{3-4x}{5x-2}.$

13–15. Restrict the domain, then find the inverse

These are quadratic functions, so they are not one-to-one on all real numbers. Choose a domain where the function is non-decreasing, usually the right half-line if the parabola opens upward.

• 13. $f(x)=(x+7)^2$
  A suitable domain is $[ -7,\infty )$.
  Then

$y=(x+7)^2 \Rightarrow x+7=\sqrt{y}$

because on this domain $x+7\ge 0$. So

$f^{-1}(x)=\sqrt{x}-7.$

• 14. $f(x)=(x-6)^2$
  A suitable domain is $[6,\infty )$.
  Then

$y=(x-6)^2 \Rightarrow x-6=\sqrt{y}$

so

$f^{-1}(x)=\sqrt{x}+6.$

• 15. $f(x)=x^2-5$
  A suitable domain is $[0,\infty )$.
  Then

$y=x^2-5 \Rightarrow x=\sqrt{y+5}$

so

$f^{-1}(x)=\sqrt{x+5}.$

16. Composition of $f(x)=\frac{7}{2x}$ and $g(x)=\frac{2}{7x}$

(a) Find $f(g(x))$ and $g(f(x))$

$f(g(x))=f\left(\frac{2}{7x}\right)=\frac{7}{2\cdot \frac{2}{7x}}=\frac{7}{\frac{4}{7x}}=\frac{49x}{4}$

That result shows the functions as written do not compose to the identity.

If the intended pair is inverse functions, then one of the expressions may have been copied incorrectly. For inverse functions, both compositions must simplify to $x$.

(b) Relationship

If two functions are inverses, then

$f(g(x))=x \quad \text{and} \quad g(f(x))=x.$

That means each function undoes the effect of the other.

17–18. Verify inverse functions by composition

• 17. $f(x)=\sqrt[3]{x-1}$ and $g(x)=x^3+1$

$f(g(x))=\sqrt[3]{(x^3+1)-1}=\sqrt[3]{x^3}=x$

$g(f(x))=(\sqrt[3]{x-1})^3+1=x-1+1=x$

So they are inverses.

• 18. $f(x)=-3x+5$ and $g(x)=\frac{x-5}{-3}$

$f(g(x))=-3\left(\frac{x-5}{-3}\right)+5=x-5+5=x$

$g(f(x))=\frac{(-3x+5)-5}{-3}=\frac{-3x}{-3}=x$

So they are inverses.

19–22. One-to-one checks from graphs

A function is one-to-one if it passes the horizontal line test.

• $f(x)=\sqrt{x}$: one-to-one
• $f(x)=\sqrt[3]{3x+1}$: one-to-one
• $f(x)=-5x+1$: one-to-one
• $f(x)=x^3-27$: one-to-one

23–24. Graph-based one-to-one test

For the graph problems, apply the horizontal line test:

• If every horizontal line hits the graph at most once, the graph is one-to-one.
• If some horizontal line hits it more than once, it is not one-to-one.

A fast check is to compare whether the graph is strictly increasing or strictly decreasing over its domain.

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Pitfalls the pros know 👇 A common mistake is to forget domain restrictions when finding inverses of quadratic functions. If you do not restrict the domain, the inverse may not be a function. Another frequent error is skipping the final step of checking by composition: for inverse functions, both $f(g(x))$ and $g(f(x))$ must simplify to $x$.

What if the problem changes? If the domain is changed, the inverse can change too. For example, a parabola restricted to the left branch uses the negative square root instead of the positive one. Also, if a rational function has a restricted domain or range, the inverse may need its own exclusions, such as values that make a denominator zero.

Tags: inverse function, one-to-one function, function composition