Question

what is the end behavior of the graph of the exponential function $f(x)=b^x$ when $0 < b < 1$?
(1 point)

• $f(x)\to\infty$ as $x\to\infty$, $f(x)\to 0$ as $x\to -\infty$
• $f(x)\to 0$ as $x\to\infty$, $f(x)\to\infty$ as $x\to -\infty$
• $f(x)\to\infty$ as $x\to\infty$, $f(x)\to -\infty$ as $x\to 0$
• $f(x)\to\infty$ as $x\to 0$, $f(x)\to -\infty$ as $x\to -\infty$

Explanation:

Step1: Recall exponential function properties

For an exponential function \( f(x) = b^x \) where \( 0 < b < 1 \), it is a decreasing function.

Step2: Analyze as \( x \to \infty \)

As \( x \) approaches positive infinity (\( x \to \infty \)), since \( b \) is between 0 and 1, raising it to a larger and larger positive power makes the value get closer to 0. So \( f(x) \to 0 \) as \( x \to \infty \).

Step3: Analyze as \( x \to -\infty \)

As \( x \) approaches negative infinity (\( x \to -\infty \)), we can rewrite \( b^x=\frac{1}{b^{-x}} \). Since \( x\to -\infty \), then \( -x\to \infty \), and \( b^{-x}=( \frac{1}{b})^x \) (wait, actually \( b^x=\frac{1}{b^{|x|}} \) when \( x\) is negative? No, more accurately, if \( x\to -\infty \), let \( x=-n \) where \( n\to \infty \). Then \( b^x = b^{-n}=\frac{1}{b^{n}} \). But since \( 0 < b < 1 \), \( \frac{1}{b}>1 \), so \( (\frac{1}{b})^n \to \infty \) as \( n\to \infty \). So \( b^x \to \infty \) as \( x\to -\infty \).

Answer:

\( f(x) \to 0 \) as \( x \to \infty \), \( f(x) \to \infty \) as \( x \to -\infty \) (the second option: \( f(x)\to 0 \) as \( x \to \infty, f(x)\to \infty \) as \( x \to -\infty \))