Let's analyze the piecewise function [tex]\( f(x) \)[/tex] for different intervals of [tex]\( x \)[/tex].
1. For [tex]\( x \leq 0 \)[/tex]:
[tex]\[
f(x) = e^{-x}
\][/tex]
This part of the function is an exponential decay function. When [tex]\( x \)[/tex] is negative or zero:
- As [tex]\( x \to -\infty \)[/tex], [tex]\( e^{-x} \to \infty \)[/tex].
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = e^{-0} = 1 \)[/tex].
2. For [tex]\( x > 0 \)[/tex]:
[tex]\[
f(x) = \frac{x^2 + 5x + 6}{x^2 + 2x - 3}
\][/tex]
We can simplify the rational function by factoring both the numerator and the denominator:
The numerator [tex]\( x^2 + 5x + 6 \)[/tex] factors to [tex]\((x + 2)(x + 3)\)[/tex].
The denominator [tex]\( x^2 + 2x - 3 \)[/tex] factors to [tex]\((x + 3)(x - 1)\)[/tex].
So, [tex]\( f(x) \)[/tex] for [tex]\( x > 0 \)[/tex] simplifies to:
[tex]\[
f(x) = \frac{(x+2)(x+3)}{(x+3)(x-1)} = \frac{x+2}{x-1}
\][/tex]
However, we need to consider that the factor [tex]\( (x + 3) \)[/tex] gets canceled out, provided [tex]\( x \neq -3 \)[/tex].
Thus, for [tex]\( x > 0 \)[/tex]:
[tex]\[
f(x) = \frac{x + 2}{x - 1}
\][/tex]
Let's consider the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \to 0^+ \)[/tex] and [tex]\( x \to \infty \)[/tex]:
- As [tex]\( x \to 0^+ \)[/tex]:
[tex]\[
f(x) = \frac{0 + 2}{0 - 1} = \frac{2}{-1} = -2
\][/tex]
- As [tex]\( x \to \infty \)[/tex], the leading terms dominate, and [tex]\( f(x) \)[/tex] approaches:
[tex]\[
f(x) \approx \frac{x}{x} = 1
\][/tex]
Thus, our piecewise function can be written canonically as:
[tex]\[
f(x) =
\begin{cases}
e^{-x} & \text { for } x \leq 0 \\
\frac{x + 2}{x - 1} & \text { for } x > 0
\end{cases}
\][/tex]
To determine the characteristics of the graph:
- For [tex]\( x \leq 0 \)[/tex], the graph will depict an exponential decay starting from [tex]\( (0,1) \)[/tex] and increasing steeply as [tex]\( x \)[/tex] goes negative.
- For [tex]\( x > 0 \)[/tex], the graph depicts a rational function with a vertical asymptote at [tex]\( x = 1 \)[/tex] and a hole at [tex]\( x = -3 \)[/tex] (not relevant since [tex]\( x > 0 \)[/tex]).
An important feature:
- The function is discontinuous at [tex]\( x = 0 \)[/tex] since there is a jump from [tex]\( f(0) = 1 \)[/tex] to [tex]\( \lim_{x \to 0^+} f(x) = -2 \)[/tex].
From the analysis, the correct graph is a combination of:
- An exponential curve decaying for [tex]\( x \leq 0 \)[/tex].
- A hyperbola-like curve approaching asymptotes for [tex]\( x > 0 \)[/tex] with a discontinuity at [tex]\( x = 0 \)[/tex].
Match this description to the options provided for your specific question to find the correct graph of [tex]\( f(x) \)[/tex].
