Composition of Functions and Invertible Functions

Introduction to Composition

Composition is the process of applying one function to the result of another. It's like an assembly line 🏭 where the output of one machine becomes the input for the next.

If we have two functions, $f$ and $g$, the composition of $f$ with $g$ (denoted as $f \circ g$) is a new function that maps an element from the domain of $g$ directly to an element in the codomain of $f$.

1. Composition of Functions

Let $f: A \to B$ and $g: B \to C$ be two functions. The composition of f and g, denoted by $g \circ f$ (read as "g composed with f" or "g of f"), is a new function that maps an element from set A directly to an element in set C.

$$(g \circ f)(x) = g(f(x)), \quad \text{for all } x \in A$$

• How it works: To find $(g \circ f)(x)$, you first compute the inner function $f(x)$, and then you use that result as the input for the outer function $g$.

• Condition for Existence: For the composition $g \circ f$ to be defined, the range of f must be a subset of the domain of g. In our factory analogy, this means the parts produced by the first machine ($f$) must fit into the second machine ($g$).

• Non-Commutative: The order of composition matters. In general, $f \circ g \ne g \circ f$.

Example: Let $f(x) = x^2$ and $g(x) = 2x+3$.

• $(g \circ f)(x) = g(f(x)) = g(x^2) = 2(x^2)+3 = \mathbf{2x^2+3}$.
• $(f \circ g)(x) = f(g(x)) = f(2x+3) = (2x+3)^2 = \mathbf{4x^2+12x+9}$.

As you can see, $(g \circ f)(x) \ne (f \circ g)(x)$.

---

2. Invertible Functions

A function is invertible if we can define an "undo" operation that takes any output and maps it back to its original input. This "undo" function is called the inverse, denoted by $f^{-1}$.

$$f(x) = y \iff f^{-1}(y) = x$$

• Condition for Invertibility: A function $f: A \to B$ is invertible if and only if it is a bijective function (i.e., it must be both one-one and onto).

  • Why One-one? For the inverse to be a function, each output must map back to a single, unique input. If the original function were many-one, its inverse would have to map one input to multiple outputs, which violates the definition of a function.
  • Why Onto? For the inverse to be defined for every element in its domain (which is the codomain of $f$), the original function's range must be equal to its codomain. If $f$ is 'into', there would be elements in the inverse's domain with nowhere to map to.

• Properties of Inverse Functions:

  • If $f$ is invertible, then $f^{-1}$ is also bijective.
  • The domain of $f^{-1}$ is the range (codomain) of $f$.
  • The range of $f^{-1}$ is the domain of $f$.
  • $(f^{-1})^{-1} = f$.
  • $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ (the "socks and shoes" rule - you undo the last operation first).

• Graphical Property: The graph of $y=f^{-1}(x)$ is the reflection of the graph of $y=f(x)$ across the line $y=x$.

---

3. How to Find the Inverse of a Function ↔️

To find the inverse of a function $f(x)$ algebraically:

1. Check for Bijectivity: First, verify that the function is both one-one and onto in its given domain and codomain.
2. Replace f(x) with y: Write the equation as $y = f(x)$.
3. Swap x and y: Interchange the variables $x$ and $y$ in the equation. This is the key step that represents the inversion.
4. Solve for y: Make $y$ the subject of the new equation. This new expression for $y$ is the inverse function, $f^{-1}(x)$.

Example: Find the inverse of the bijective function $f(x) = \frac{2x+1}{x-3}$.

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

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

3. Solve for y:

   $x(y-3) = 2y+1$

   $xy - 3x = 2y+1$

   $xy - 2y = 3x+1$

   $y(x - 2) = 3x+1$

   $y = \frac{3x+1}{x-2}$.

4. State the Inverse: $f^{-1}(x) = \frac{3x+1}{x-2}$.

Example 1: Finding Compositions

Question: Let $f(x) = |x|$ and $g(x) = [x]$ (Greatest Integer Function). Find $(g \circ f)(-2.5)$ and $(f \circ g)(3.7)$.

Explanation: Function composition $(g \circ f)(x)$ means we apply the inner function $f$ first, and then apply the outer function $g$ to the result.

Part 1: Find $(g \circ f)(-2.5)$

1. Evaluate the inner function: First, find $f(-2.5)$. $f(-2.5) = |-2.5| = 2.5$.

2. Evaluate the outer function: Now, use the result from step 1 as the input for $g$.

   $(g \circ f)(-2.5) = g(f(-2.5)) = g(2.5) = [2.5] = 2$.

Part 2: Find $(f \circ g)(3.7)$

1. Evaluate the inner function: First, find $g(3.7)$.

   $g(3.7) = [3.7] = 3$.

2. Evaluate the outer function: Now, use the result from step 1 as the input for $f$.

   $(f \circ g)(3.7) = f(g(3.7)) = f(3) = |3| = 3$.

Example 2: Proving a Function is Bijective

Question: Show that the function $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x)=4x+3$ is bijective.

Explanation: A function is bijective if it is both one-one (injective) and onto (surjective).

1. Check for One-one (Injective): We must show that if $f(x_1) = f(x_2)$, then $x_1 = x_2$.

• Assume $f(x_1) = f(x_2)$.
• $4x_1+3 = 4x_2+3$
• $4x_1 = 4x_2$
• $x_1 = x_2$. Since the condition holds, the function is one-one.

2. Check for Onto (Surjective): We must show that for any $y$ in the codomain $\mathbb{R}$, there exists a pre-image $x$ in the domain $\mathbb{R}$ such that $f(x)=y$.

• Let $y = 4x+3$.
• Solve for $x$: $y-3 = 4x \implies x = \frac{y-3}{4}$.
• For any real number $y$, the expression $\frac{y-3}{4}$ produces a valid real number $x$. Thus, every element in the codomain has a pre-image.
• The function is onto.

Conclusion: Since f is both one-one and onto, it is bijective.

Example 3: Finding the Inverse of a Linear Function

Question: Find the inverse of the function $f(x)=4x+3$.

Explanation: From the previous example, we know this function is bijective, so an inverse exists.

1. Replace $f(x)$ with y: $y = 4x+3$.
2. Swap the variables x and y: $x = 4y+3$.
3. Solve for the new y: $x-3 = 4y$ $y = \frac{x-3}{4}$.
4. State the inverse function: The new expression for y is the inverse function, $f^{-1}(x)$. $$f^{-1}(x) = \frac{x-3}{4}$$

Example 4: Finding the Inverse of a Rational Function

Question: Let $f: \mathbb{R}-\{2\} \to \mathbb{R}-\{1\}$ be defined by $f(x) = \frac{x-1}{x-2}$. Find its inverse.

Explanation: We assume the function is bijective as defined by its specific domain and codomain.

1. Replace $f(x)$ with y: $y = \frac{x-1}{x-2}$.
2. Swap x and y: $x = \frac{y-1}{y-2}$.
3. Solve for y:
   • $x(y-2) = y-1$
   • $xy - 2x = y-1$
   • $xy - y = 2x - 1$
   • $y(x-1) = 2x - 1$
   • $y = \frac{2x-1}{x-1}$.
4. State the inverse function: $$f^{-1}(x) = \frac{2x-1}{x-1}$$

Example 5: Composition of a Function and its Inverse

Question: Let $f(x)=4x+3$. Verify that $(f \circ f^{-1})(x) = x$.

Explanation: An important property of inverse functions is that composing a function with its inverse (in either order) results in the identity function, $I(x)=x$.

1. Recall the Functions:

   • $f(x) = 4x+3$
   • From Example 3, we know $f^{-1}(x) = \frac{x-3}{4}$.

2. Perform the Composition: $$(f \circ f^{-1})(x) = f(f^{-1}(x))$$ This means we substitute the entire expression for $f^{-1}(x)$ into the 'x' of $f(x)$.

   $f\left(\frac{x-3}{4}\right) = 4\left(\frac{x-3}{4}\right) + 3$.

3. Simplify: The '4's cancel out: $(x-3) + 3 = x$.

Conclusion: We have verified that $(f \circ f^{-1})(x) = x$.

Example 6: Finding Inverse of a Quadratic Function

Question: Let $f: [0, \infty) \to [4, \infty)$ be a function defined by $f(x)=x^2+4$. Find the inverse of f.

Explanation: 1. Check Bijectivity:

• The standard function $f(x)=x^2$ is many-one. However, by restricting the domain to $[0, \infty)$, we ensure that no two x-values produce the same y-value, making it one-one.
• The range of $x^2+4$ on this domain is $[4, \infty)$, which is equal to the given codomain. This makes the function onto.

Since the function is bijective on the given domain/codomain, an inverse exists.

2. Find the Inverse:

• Set y=f(x): $y = x^2+4$.
• Swap x and y: $x = y^2+4$.
• Solve for y: $y^2 = x-4 \implies y = \pm\sqrt{x-4}$.

3. Choose the Correct Root: The range of the inverse function, $f^{-1}$, must be the domain of the original function, $f$. The domain of $f$ was given as $[0, \infty)$. Therefore, the output of our inverse function ($y$) must be non-negative. To ensure this, we must choose the positive root.

$y = +\sqrt{x-4}$.

Answer: $f^{-1}(x) = \sqrt{x-4}$.

Example 6: Finding Inverse of a Quadratic Function

Question: Let $f: [0, \infty) \to [4, \infty)$ be a function defined by $f(x)=x^2+4$. Find the inverse of f.

Explanation: 1. Check Bijectivity:

• The standard function $f(x)=x^2$ is many-one. However, by restricting the domain to $[0, \infty)$, we ensure that no two x-values produce the same y-value, making it one-one.
• The range of $x^2+4$ on this domain is $[4, \infty)$, which is equal to the given codomain. This makes the function onto. Since the function is bijective on the given domain/codomain, an inverse exists.

2. Find the Inverse:

• Set y=f(x): $y = x^2+4$.
• Swap x and y: $x = y^2+4$.
• Solve for y: $y^2 = x-4 \implies y = \pm\sqrt{x-4}$.

3. Choose the Correct Root: The range of the inverse function, $f^{-1}$, must be the domain of the original function, $f$. The domain of $f$ was given as $[0, \infty)$. Therefore, the output of our inverse function ($y$) must be non-negative. To ensure this, we must choose the positive root. $y = +\sqrt{x-4}$.

Answer: $f^{-1}(x) = \sqrt{x-4}$.

Example 7: Composition with Itself

Question: If $f(x) = \frac{1}{1-x}$, find $(f \circ f \circ f)(x)$.

Explanation: We solve this by composing the function in stages.

Stage 1: Calculate $(f \circ f)(x)$

$(f \circ f)(x) = f(f(x)) = f\left(\frac{1}{1-x}\right)$.

Substitute this into the function's definition: $$f\left(\frac{1}{1-x}\right) = \frac{1}{1 - \frac{1}{1-x}} = \frac{1}{\frac{(1-x)-1}{1-x}} = \frac{1-x}{-x} = \frac{x-1}{x}$$

Stage 2: Calculate $(f \circ f \circ f)(x)$

This is the same as $f( (f \circ f)(x) )$. We use the result from Stage 1 as the input.

$$f\left(\frac{x-1}{x}\right) = \frac{1}{1 - \frac{x-1}{x}} = \frac{1}{\frac{x-(x-1)}{x}} = \frac{x}{x-x+1} = \frac{x}{1} = x$$

Answer: $(f \circ f \circ f)(x) = x$.

Example 8: Finding Domain of a Composition

Question: Let $f(x) = \sqrt{x}$ and $g(x) = x-2$. Find the domain of $f \circ g$.

Explanation: 1. Define the Composite Function:

$(f \circ g)(x) = f(g(x)) = f(x-2) = \sqrt{x-2}$.

2. Determine the Condition for the Domain: The expression $\sqrt{x-2}$ is defined in the real numbers only if the quantity inside the square root is non-negative.

Condition: $x-2 \ge 0$.

3. Solve the Inequality: $x \ge 2$.

Formal Approach: The domain of $f \circ g$ is the set of all $x$ in the domain of $g$ such that the output $g(x)$ is in the domain of $f$.

• Domain of $g(x)=x-2$ is $\mathbb{R}$.
• Domain of $f(x)=\sqrt{x}$ is $[0, \infty)$.
• We need the output of $g$, which is $x-2$, to be in the domain of $f$. So, $x-2 \in [0, \infty)$, which means $x-2 \ge 0$. This gives $x \ge 2$.

Answer: The domain is $[2, \infty)$.

Example 9: Finding a Value in a Composition

Question: If $f(x)=x^2-1$ and $g(x)=2x+1$, find a value of x for which $(f \circ g)(x) = (g \circ f)(x)$.

Explanation: 1. Calculate $(f \circ g)(x)$:

$$(f \circ g)(x) = f(g(x)) = f(2x+1) = (2x+1)^2-1$$ $$= (4x^2+4x+1)-1 = 4x^2+4x$$

2. Calculate $(g \circ f)(x)$: $$(g \circ f)(x) = g(f(x)) = g(x^2-1) = 2(x^2-1)+1$$ $$= 2x^2-2+1 = 2x^2-1$$

3. Set the Expressions Equal and Solve: $$4x^2+4x = 2x^2-1$$ $$2x^2+4x+1 = 0$$ This is a quadratic equation. We use the quadratic formula $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ Here, $a=2, b=4, c=1$. $$x = \frac{-4 \pm \sqrt{4^2-4(2)(1)}}{2(2)} = \frac{-4 \pm \sqrt{16-8}}{4} = \frac{-4 \pm \sqrt{8}}{4} = \frac{-4 \pm 2\sqrt{2}}{4}$$ $x = -1 \pm \frac{\sqrt{2}}{2}$.

Answer: Two possible values are $-1 + \frac{\sqrt{2}}{2}$ and $-1 - \frac{\sqrt{2}}{2}$.

Example 10: Inverse of a Bijective Piecewise Function

Question: Find the inverse of the function $f(x) = \begin{cases} x, & x < 1 \\ x^2, & 1 \le x \le 4 \end{cases}$.

Explanation: We find the inverse for each piece separately, paying close attention to the domains and ranges.

1. Analyze the First Piece:

• Function: $y=x$ for the domain $(-\infty, 1)$.
• Range: For this domain, the range is also $(-\infty, 1)$.
• Inverse: Swapping $x$ and $y$ gives $x=y$, so $f^{-1}(x)=x$. The domain of this inverse piece is the range of the original piece, so this is valid for $x < 1$.

2. Analyze the Second Piece:

• Function: $y=x^2$ for the domain $[1, 4]$.
• Range: For this domain, the range is $[1^2, 4^2] = [1, 16]$.
• Inverse: Swapping $x$ and $y$ gives $x=y^2$, so $y=\pm\sqrt{x}$. Since the range of the inverse must be the domain of the original ($[1,4]$), we choose the positive root. So, $f^{-1}(x)=\sqrt{x}$. The domain of this inverse piece is the range of the original piece, so this is valid for $1 \le x \le 16$.

3. Combine the Inverse Pieces: We combine the results from each part to define the complete inverse function.

Answer: $$f^{-1}(x) = \begin{cases} x, & x < 1 \\ \sqrt{x}, & 1 \le x \le 16 \end{cases}$$