For the following functions f and g, determine with reason whether (fog)(x) exists: (a) f : R → R, where f(x) = x 3 and g : [− 3 2 , ∞) → R, with g(x) = √ 2x + 3. (b) f : R → R, where f(x) = 5 x−6 and g : [ln(5), ∞) → R, with g(x) = e x .

To determine whether the composition of functions \( (f \circ g)(x) \) exists, we need to ensure that the output of the function \( g(x) \) lies within the domain of the function \( f(x) \). Specifically, \( g(x) \) must produce values that are within the domain of \( f \).

Let's analyze each case:

### (a) \( f(x) = x^3 \) and \( g(x) = \sqrt{2x + 3} \)

- **Function \( f(x) = x^3 \)**:
  - Domain: \( f : \mathbb{R} \to \mathbb{R} \). This means \( f(x) \) is defined for all real numbers \( x \).

- **Function \( g(x) = \sqrt{2x + 3} \)**:
  - Domain: \( g : \left[-\frac{3}{2}, \infty\right) \to \mathbb{R} \).
  - This domain is derived because the expression inside the square root, \( 2x + 3 \), must be non-negative:
    \[
    2x + 3 \geq 0
    \]
    \[
    x \geq -\frac{3}{2}
    \]
  - Therefore, \( g(x) \) is defined for \( x \geq -\frac{3}{2} \).

- **Composition \( (f \circ g)(x) \)**:
  - For \( (f \circ g)(x) = f(g(x)) \) to exist, the output of \( g(x) \) must lie within the domain of \( f(x) \).
  - Since \( f(x) = x^3 \) is defined for all real numbers, there are no restrictions on the values that \( g(x) \) can take.
  - The function \( g(x) = \sqrt{2x + 3} \) always produces non-negative real numbers (since the square root of a non-negative number is non-negative).

**Conclusion for (a):**  
\( (f \circ g)(x) \) exists for all \( x \in \left[-\frac{3}{2}, \infty\right) \).

### (b) \( f(x) = \frac{5}{x - 6} \) and \( g(x) = e^x \)

- **Function \( f(x) = \frac{5}{x - 6} \)**:
  - Domain: \( f : \mathbb{R} \setminus \{6\} \to \mathbb{R} \). This means \( f(x) \) is defined for all real numbers \( x \) except \( x = 6 \) (since division by zero is undefined).

- **Function \( g(x) = e^x \)**:
  - Domain: \( g : [\ln(5), \infty) \to \mathbb{R} \).
  - This means \( g(x) \) is defined for all \( x \geq \ln(5) \).

- **Composition \( (f \circ g)(x) \)**:
  - For \( (f \circ g)(x) = f(g(x)) \) to exist, \( g(x) \) must lie within the domain of \( f(x) \).
  - \( g(x) = e^x \) produces values in the range \( [e^{\ln(5)}, \infty) = [5, \infty) \) since the exponential function is always positive and strictly increasing.
  - Therefore, \( g(x) \) can produce values from \( 5 \) to \( \infty \).
  - For \( f(x) = \frac{5}{x - 6} \) to be defined, \( g(x) \) should not equal \( 6 \). However, \( g(x) = e^x \) will never equal \( 6 \) for any real \( x \) (because \( e^x = 6 \) corresponds to \( x = \ln(6) \), which is not in the interval \( [\ln(5), \infty) \)).

**Conclusion for (b):**  
\( (f \circ g)(x) \) exists for all \( x \in [\ln(5), \infty) \) because \( g(x) \) does not produce the value \( 6 \), which is the only value that would make \( f(x) \) undefined.

### Summary:
- **(a) \( (f \circ g)(x) \) exists** for \( x \in \left[-\frac{3}{2}, \infty\right) \).
- **(b) \( (f \circ g)(x) \) exists** for \( x \in [\ln(5), \infty) \).

Would you like further details or have any questions?

### Related Questions
1. What are the general conditions for the existence of a composition of functions?
2. How does the range of one function affect the domain of its composition?
3. Can we have compositions where the domain of one function restricts the composition?
4. How can the inverse function of a composition be determined?
5. What are the common pitfalls to avoid when working with compositions of functions?

### Tip
When determining whether a composition of functions exists, always check that the output of the inner function lies within the domain of the outer function. This ensures that the composition is well-defined.

Explore the conditions for the existence of the composition (f ∘ g)(x) where f(x) = x^3 and g(x) = sqrt(2x + 3). Understand how the domain of g(x) influences the domain of f(x). Suitable for advanced high school students and beyond.

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