Prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using Venn diagrams. If A = {1, 2, 3, 4} and B = {2, 3, 4, 5}, find A ∪ B, A ∩ B, and A - B. Find the domain and range of the function f(... Prove that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) using Venn diagrams. If A = {1, 2, 3, 4} and B = {2, 3, 4, 5}, find A ∪ B, A ∩ B, and A - B. Find the domain and range of the function f(x) = √(x - 3). If f(x) = (2x + 3) / (x - 1), find f^(-1)(x). Prove that if f and g are bijective, then f ∘ g is also bijective. Read more

Steps to Solve

1. Proving the Set Equality Using Venn Diagrams

To prove $A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)$ using Venn diagrams, we can visualize each side:

• Left Side: $A ∪ (B ∩ C)$ includes all elements in $A$ and elements that are both in $B$ and $C$.
• Right Side: $(A ∪ B) ∩ (A ∪ C)$ includes elements in either $A$ or $B$, and simultaneously in either $A$ or $C$.

By overlapping parts of the circles representing each set, we can see that both sides end up covering the same areas, thus proving the equality.

2. Finding Set Operations

Given:

• $A = {1, 2, 3, 4}$
• $B = {2, 3, 4, 5}$

Now, calculate:

• Union: $A ∪ B = {1, 2, 3, 4, 5}$

• Intersection: $A ∩ B = {2, 3, 4}$

• Difference: $A - B = {1}$

3. Finding Domain and Range of the Function

For the function $f(x) = \sqrt{x - 3}$:

• Domain: Set the expression inside the square root to be non-negative:

  $$x - 3 ≥ 0 \Rightarrow x ≥ 3$$

  Thus, the domain is $[3, ∞)$.

• Range: For $y = f(x)$, the smallest value that $f(x)$ can take is $0$ (when $x = 3$), so the range is $[0, ∞)$.

4. Finding the Inverse Function

For $f(x) = \frac{2x + 3}{x - 1}$:

• Start by replacing $f(x)$ with $y$:

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

• Swap $x$ and $y$ and solve for $y$:

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

  Multiply both sides by $(y - 1)$:

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

  Rearranging gives:

  $$xy - x = 2y + 3$$

  $$xy - 2y = x + 3$$

  Factor out $y$:

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

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

Thus, the inverse function is $f^{-1}(x) = \frac{x + 3}{x - 2}$.

5. Proving Composition of Bijective Functions

To prove that if $f$ and $g$ are bijective, then $f ∘ g$ is also bijective:

• Both $f$ and $g$ have a one-to-one correspondence (bijective).
• Thus, for every $x$ in the domain of $g$, there exists a unique $y$ in the range of $g$, and for each corresponding $y$, there’s a unique $z$ in the range of $f$.
• Therefore, $f ∘ g$ is also one-to-one, and every element in the target set has a unique pre-image from the domain, confirming that $f ∘ g$ is bijective.