Question

1. which statement about ( f(x) = 3^x - 6 ) is true? circle the correct letter.

a. ( f(x) ) has both an absolute minimum and an absolute maximum.
b. ( f(x) ) has an absolute minimum.
c. ( f(x) ) has an absolute maximum.
d. ( f(x) ) has neither an absolute minimum nor an absolute maximum.

Explanation:

Step1: Analyze the function \( f(x) = 3^x - 6 \)

The function \( y = 3^x \) is an exponential function with base \( 3>1 \), so it is an increasing function. Its range is \( (0, +\infty) \) because as \( x \to -\infty \), \( 3^x \to 0 \), and as \( x \to +\infty \), \( 3^x \to +\infty \).

Step2: Analyze the range of \( f(x) = 3^x - 6 \)

For \( f(x)=3^x - 6 \), when \( x \to -\infty \), \( 3^x \to 0 \), so \( f(x) \to 0 - 6=-6 \). But since \( 3^x>0 \) for all real \( x \), \( f(x)=3^x - 6>-6 \) for all real \( x \). When \( x \to +\infty \), \( 3^x \to +\infty \), so \( f(x) \to +\infty \).

Step3: Determine absolute extrema

• Absolute minimum: An absolute minimum would be a value \( m \) such that \( f(x)\geq m \) for all \( x \) and there exists some \( x_0 \) with \( f(x_0)=m \). But as \( x \to -\infty \), \( f(x) \to -6 \) but never actually reaches \( -6 \) (since \( 3^x>0 \)), so there is no absolute minimum.
• Absolute maximum: An absolute maximum would be a value \( M \) such that \( f(x)\leq M \) for all \( x \) and there exists some \( x_1 \) with \( f(x_1)=M \). But as \( x \to +\infty \), \( f(x) \to +\infty \), so there is no upper bound, hence no absolute maximum.

Answer:

d. \( f(x) \) has neither an absolute minimum nor an absolute maximum.