Answer :
Let's carefully analyze the given piecewise function:
[tex]\[ f(x) = \begin{cases} e^{-x} & \text{for } x \leq 0 \\ \frac{x^2 + x - 6}{x^2 + 2x - 3} & \text{for } x > 0 \end{cases} \][/tex]
### Analysis for [tex]\( x \leq 0 \)[/tex]:
For [tex]\( x \leq 0 \)[/tex], the function is [tex]\( f(x) = e^{-x} \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(0) = e^0 = 1 \)[/tex].
- As [tex]\( x \)[/tex] becomes more negative, [tex]\( e^{-x} \)[/tex] grows exponentially. For example:
- When [tex]\( x = -1 \)[/tex], [tex]\( f(-1) = e^1 \approx 2.718 \)[/tex].
- When [tex]\( x = -2 \)[/tex], [tex]\( f(-2) = e^2 \approx 7.389 \)[/tex].
In summary, [tex]\( f(x) \)[/tex] for [tex]\( x \leq 0 \)[/tex] is an exponential function that increases rapidly as [tex]\( x \)[/tex] becomes more negative.
### Analysis for [tex]\( x > 0 \)[/tex]:
For [tex]\( x > 0 \)[/tex], the function is [tex]\( f(x) = \frac{x^2 + x - 6}{x^2 + 2x - 3} \)[/tex].
First, let's simplify the rational function:
[tex]\[ f(x) = \frac{x^2 + x - 6}{x^2 + 2x - 3} \][/tex]
Factor both the numerator and the denominator:
- Numerator: [tex]\( x^2 + x - 6 = (x + 3)(x - 2) \)[/tex]
- Denominator: [tex]\( x^2 + 2x - 3 = (x + 3)(x - 1) \)[/tex]
Thus, the simplified form is:
[tex]\[ f(x) = \frac{(x + 3)(x - 2)}{(x + 3)(x - 1)} \][/tex]
As long as [tex]\( x \neq -3 \)[/tex], this simplifies to:
[tex]\[ f(x) = \frac{x - 2}{x - 1} \][/tex]
This rational function has a vertical asymptote at [tex]\( x = 1 \)[/tex] and a removable discontinuity at [tex]\( x = -3 \)[/tex], which is not relevant for [tex]\( x > 0 \)[/tex].
Key points and behaviors:
- For [tex]\( x \)[/tex] approaching 1 from the right, the function [tex]\( f(x) \)[/tex] grows negatively large.
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to 1 \)[/tex] (since the degree of the numerator and the denominator is the same).
Quick values:
- When [tex]\( x = 2 \)[/tex], [tex]\( f(2) = \frac{2 - 2}{2 - 1} = 0 \)[/tex].
- For [tex]\( x \)[/tex] slightly greater than 1:
- When [tex]\( x = 1.1 \)[/tex], [tex]\( f(1.1) = \frac{1.1 - 2}{1.1 - 1} = \frac{-0.9}{0.1} = -9 \)[/tex].
### Graph Characteristics Summary:
- For [tex]\( x \leq 0 \)[/tex]: Exponential decay that increases rapidly as [tex]\( x \)[/tex] decreases.
- For [tex]\( x > 0 \)[/tex]: A rational function with a vertical asymptote at [tex]\( x = 1 \)[/tex] and crosses [tex]\( y = 0 \)[/tex] at [tex]\( x = 2 \)[/tex]. The values approach [tex]\( y = 1 \)[/tex] as [tex]\( x \)[/tex] becomes very large.
Given the above analysis, the correct graph of the piecewise function will have:
- An increasing exponential curve for negative [tex]\( x \)[/tex].
- A rational function for positive [tex]\( x \)[/tex] with a vertical asymptote at [tex]\( x = 1 \)[/tex], a point at (2, 0), and asymptotically approaching [tex]\( y = 1 \)[/tex].
To select the correct graph from multiple choices, you would look for the graph that best matches these characteristics.
[tex]\[ f(x) = \begin{cases} e^{-x} & \text{for } x \leq 0 \\ \frac{x^2 + x - 6}{x^2 + 2x - 3} & \text{for } x > 0 \end{cases} \][/tex]
### Analysis for [tex]\( x \leq 0 \)[/tex]:
For [tex]\( x \leq 0 \)[/tex], the function is [tex]\( f(x) = e^{-x} \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(0) = e^0 = 1 \)[/tex].
- As [tex]\( x \)[/tex] becomes more negative, [tex]\( e^{-x} \)[/tex] grows exponentially. For example:
- When [tex]\( x = -1 \)[/tex], [tex]\( f(-1) = e^1 \approx 2.718 \)[/tex].
- When [tex]\( x = -2 \)[/tex], [tex]\( f(-2) = e^2 \approx 7.389 \)[/tex].
In summary, [tex]\( f(x) \)[/tex] for [tex]\( x \leq 0 \)[/tex] is an exponential function that increases rapidly as [tex]\( x \)[/tex] becomes more negative.
### Analysis for [tex]\( x > 0 \)[/tex]:
For [tex]\( x > 0 \)[/tex], the function is [tex]\( f(x) = \frac{x^2 + x - 6}{x^2 + 2x - 3} \)[/tex].
First, let's simplify the rational function:
[tex]\[ f(x) = \frac{x^2 + x - 6}{x^2 + 2x - 3} \][/tex]
Factor both the numerator and the denominator:
- Numerator: [tex]\( x^2 + x - 6 = (x + 3)(x - 2) \)[/tex]
- Denominator: [tex]\( x^2 + 2x - 3 = (x + 3)(x - 1) \)[/tex]
Thus, the simplified form is:
[tex]\[ f(x) = \frac{(x + 3)(x - 2)}{(x + 3)(x - 1)} \][/tex]
As long as [tex]\( x \neq -3 \)[/tex], this simplifies to:
[tex]\[ f(x) = \frac{x - 2}{x - 1} \][/tex]
This rational function has a vertical asymptote at [tex]\( x = 1 \)[/tex] and a removable discontinuity at [tex]\( x = -3 \)[/tex], which is not relevant for [tex]\( x > 0 \)[/tex].
Key points and behaviors:
- For [tex]\( x \)[/tex] approaching 1 from the right, the function [tex]\( f(x) \)[/tex] grows negatively large.
- As [tex]\( x \to \infty \)[/tex], [tex]\( f(x) \to 1 \)[/tex] (since the degree of the numerator and the denominator is the same).
Quick values:
- When [tex]\( x = 2 \)[/tex], [tex]\( f(2) = \frac{2 - 2}{2 - 1} = 0 \)[/tex].
- For [tex]\( x \)[/tex] slightly greater than 1:
- When [tex]\( x = 1.1 \)[/tex], [tex]\( f(1.1) = \frac{1.1 - 2}{1.1 - 1} = \frac{-0.9}{0.1} = -9 \)[/tex].
### Graph Characteristics Summary:
- For [tex]\( x \leq 0 \)[/tex]: Exponential decay that increases rapidly as [tex]\( x \)[/tex] decreases.
- For [tex]\( x > 0 \)[/tex]: A rational function with a vertical asymptote at [tex]\( x = 1 \)[/tex] and crosses [tex]\( y = 0 \)[/tex] at [tex]\( x = 2 \)[/tex]. The values approach [tex]\( y = 1 \)[/tex] as [tex]\( x \)[/tex] becomes very large.
Given the above analysis, the correct graph of the piecewise function will have:
- An increasing exponential curve for negative [tex]\( x \)[/tex].
- A rational function for positive [tex]\( x \)[/tex] with a vertical asymptote at [tex]\( x = 1 \)[/tex], a point at (2, 0), and asymptotically approaching [tex]\( y = 1 \)[/tex].
To select the correct graph from multiple choices, you would look for the graph that best matches these characteristics.
