`f(x)=e^x : R^+ ->R and g(x)=x^2-x : R->R` Find domain and range of fog (x) and gof(x)

Explanation

To find the domain and range of the composite functions \( f(g(x)) \) and \( g(f(x)) \), we will follow these steps:

### Step 1: Define the Functions
We have:
- \( f(x) = e^x \) where \( f: \mathbb{R}^+ \to \mathbb{R} \)
- \( g(x) = x^2 - x \) where \( g: \mathbb{R} \to \mathbb{R} \)

### Step 2: Find the Domain of \( f(g(x)) \)
To find the domain of \( f(g(x)) \), we need to ensure that \( g(x) \) is in the domain of \( f \). Since \( f \) is defined for \( \mathbb{R}^+ \) (i.e., \( g(x) > 0 \)), we need to solve the inequality:

\[ g(x) = x^2 - x > 0 \]

### Step 3: Solve the Inequality
Factoring gives us:

\[ x(x - 1) > 0 \]

To solve this inequality, we find the critical points by setting \( g(x) = 0 \):

\[ x(x - 1) = 0 \implies x = 0 \quad \text{or} \quad x = 1 \]

Now we test the intervals determined by these critical points: \( (-\infty, 0) \), \( (0, 1) \), and \( (1, \infty) \).

- For \( x < 0 \) (e.g., \( x = -1 \)): \( g(-1) = (-1)(-2) = 2 > 0 \)
- For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)): \( g(0.5) = (0.5)(-0.5) = -0.25 < 0 \)
- For \( x > 1 \) (e.g., \( x = 2 \)): \( g(2) = (2)(1) = 2 > 0 \)

Thus, the solution to the inequality \( g(x) > 0 \) is:

\[ x \in (-\infty, 0) \cup (1, \infty) \]

### Step 4: Find the Range of \( f(g(x)) \)
Next, we need to find the range of \( f(g(x)) = e^{g(x)} \). Since \( g(x) \) is a quadratic function, we find its minimum value to determine the range.

The vertex of \( g(x) = x^2 - x \) occurs at:

\[ x = -\frac{b}{2a} = \frac{1}{2} \]

Calculating \( g\left(\frac{1}{2}\right) \):

\[ g\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \]

Since \( g(x) \) has a minimum value of \( -\frac{1}{4} \) and approaches infinity as \( x \) goes to \( -\infty \) or \( +\infty \), the range of \( g(x) \) is:

\[ g(x) \in \left[-\frac{1}{4}, \infty\right) \]

Thus, the range of \( f(g(x)) = e^{g(x)} \) is:

\[ f(g(x)) \in \left[e^{-\frac{1}{4}}, \infty\right) \]

### Step 5: Find the Domain of \( g(f(x)) \)
Now, we find \( g(f(x)) \). Since \( f(x) \) is defined for all \( x \in \mathbb{R}^+ \), the domain of \( g(f(x)) \) is:

\[ x \in \mathbb{R}^+ \]

### Step 6: Find the Range of \( g(f(x)) \)
We compute \( g(f(x)) = g(e^x) = (e^x)^2 - e^x = e^{2x} - e^x \).

To find the range, we analyze the function \( h(x) = e^{2x} - e^x \).

1. As \( x \to 0 \), \( h(0) = 1 - 1 = 0 \).
2. As \( x \to \infty \), \( h(x) \to \infty \).

The function \( h(x) \) is always increasing for \( x > 0 \) since its derivative \( h'(x) = 2e^{2x} - e^x \) is positive for \( x > 0 \).

Thus, the range of \( g(f(x)) \) is:

\[ g(f(x)) \in [0, \infty) \]

### Summary of Results
- **Domain of \( f(g(x)) \)**: \( (-\infty, 0) \cup (1, \infty) \)
- **Range of \( f(g(x)) \)**: \( \left[e^{-\frac{1}{4}}, \infty\right) \)
- **Domain of \( g(f(x)) \)**: \( (0, \infty) \)
- **Range of \( g(f(x)) \)**: \( [0, \infty) \)