Question

consider the graph of the function $f(x) = 10^x$.
what is the range of function $g$ if $g(x) = -2f(x) + 1$?
\\(\bigcirc\\) a. \\((-\infty, 1)\\)
\\(\bigcirc\\) b. \\((-\infty, 2)\\)
\\(\bigcirc\\) c. \\((-2, \infty)\\)
\\(\bigcirc\\) d. \\((0, \infty)\\)

Explanation:

Step1: Find range of \( f(x) = 10^x \)

The function \( f(x)=10^x \) is an exponential function. For any real number \( x \), \( 10^x>0 \). So the range of \( f(x) \) is \( (0, \infty) \).

Step2: Analyze transformation for \( g(x)=-2f(x)+1 \)

First, multiply \( f(x) \) by -2: if \( f(x)>0 \), then \( -2f(x)<0 \) (since multiplying by a negative number reverses the inequality). Then add 1: \( -2f(x)+1 < 0 + 1=1 \). As \( f(x) \) can get arbitrarily large (approaching \( \infty \)), \( -2f(x) \) will approach \( -\infty \), so \( -2f(x)+1 \) will also approach \( -\infty \). So the range of \( g(x) \) is \( (-\infty, 1) \).

Answer:

A. \( (-\infty, 1) \)