Question

1 multiple answer 20 points which of the following is the correct end behavior of the function $f(x)=2^{x}$ as $x \to \infty, f(x) \to \infty$; as $x \to -\infty, f(x) \to 0$ as $x \to \infty, f(x) \to -\infty$; as $x \to -\infty, f(x) \to \infty$ as $x \to \infty, f(x) \to 0$; as $x \to -\infty, f(x) \to 0$ as $x \to \infty, f(x) \to 0$; as $x \to -\infty, f(x) \to \infty$

Explanation:

Step1: Analyze the behavior as \(x\to\infty\)

For the exponential function \(y = a^{x}\) (\(a>1\), here \(a = 2\)), when \(x\to\infty\), \(y=2^{x}\to\infty\). Mathematically, \(\lim_{x\to\infty}2^{x}=\infty\) since as \(x\) gets larger and larger, \(2^{x}\) grows without bound.

Step2: Analyze the behavior as \(x\to-\infty\)

When \(x\to-\infty\), we can rewrite \(y = 2^{x}=\frac{1}{2^{-x}}\). Let \(t=-x\), then as \(x\to-\infty\), \(t\to\infty\). So \(y=\frac{1}{2^{t}}\), and \(\lim_{t\to\infty}\frac{1}{2^{t}} = 0\). So \(\lim_{x\to-\infty}2^{x}=0\)

Answer:

As \(x\to\infty,f(x)\to\infty\); as \(x\to-\infty,f(x)\to0\)