Question

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

Explanation:

Step1: Find range of \( f(x) \)

The function \( f(x) = 10^x \) is an exponential function. For exponential functions of the form \( a^x \) (where \( a>1 \)), the range is \( (0, \infty) \). So, \( f(x)>0 \).

Step2: Analyze transformation for \( g(x) \)

We have \( g(x)= -2f(x)+1 \). First, multiply \( f(x) \) by -2. When we multiply a positive number (since \( f(x)>0 \)) by -2, the inequality reverses: \( -2f(x) < 0 \) (because multiplying by a negative number reverses the inequality sign).

Step3: Add 1 to the inequality

Now, add 1 to both sides of \( -2f(x) < 0 \). We get \( -2f(x)+1 < 0 + 1 \), which simplifies to \( g(x) < 1 \). So the range of \( g(x) \) is \( (-\infty, 1) \)? Wait, no, wait. Wait, let's re - do the inequality steps carefully.

Wait, \( f(x)>0 \). Multiply both sides by - 2: when we multiply an inequality by a negative number, the direction of the inequality sign changes. So \( - 2f(x)<-2\times0 = 0 \). Then add 1 to both sides: \( -2f(x)+1 < 0 + 1\), so \( g(x)<1 \). But also, as \( f(x) \) can get arbitrarily large (since \( 10^x \) goes to \( \infty \) as \( x ightarrow\infty \)), \( -2f(x) \) will go to \( -\infty \), and then \( -2f(x)+1 \) will also go to \( -\infty \). So the range of \( g(x) \) is \( (-\infty, 1) \)? Wait, no, let's check again.

Wait, \( f(x)=10^x \), range \( (0, \infty) \). Let \( y = f(x) \), so \( y\in(0, \infty) \). Then \( g(x)=-2y + 1 \). Let's find the range of \( -2y+1 \) where \( y>0 \).

Solve for the range of \( z=-2y + 1 \), \( y > 0 \).

We can rewrite \( z=-2y + 1\) as \( y=\frac{1 - z}{2} \). Since \( y>0 \), we have \( \frac{1 - z}{2}>0 \). Multiply both sides by 2 (positive, so inequality sign remains): \( 1 - z>0 \), then \( -z>-1 \), multiply both sides by - 1 (inequality sign reverses): \( z < 1 \). And as \( y ightarrow0^+ \), \( z=-2\times0 + 1=1 \) (but \( y \) never actually reaches 0, so \( z \) never actually reaches 1). As \( y ightarrow\infty \), \( z=-2\times\infty+1=-\infty \). So the range of \( g(x) \) is \( (-\infty, 1) \)? Wait, but the options have D as \( (-\infty, 1) \)? Wait, no, wait in the options, option D is \( (-\infty, 1) \)? Wait, no, let's check the options again.

Wait the options are:

A. \( (-2, \infty) \)

B. \( (-\infty, 2) \)

C. \( (0, \infty) \)

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

Wait, I think I made a mistake in the inequality steps. Let's start over.

1. Range of \( f(x)=10^x \): For any real number \( x \), \( 10^x>0 \). So \( f(x)\in(0, \infty) \).

2. For \( g(x)=-2f(x)+1 \), let's perform the transformations on the range.

First, multiply \( f(x) \) by - 2: If \( f(x)>0 \), then \( -2f(x)<0 \) (because multiplying a positive number by a negative number gives a negative number, and as \( f(x) \) increases, \( -2f(x) \) decreases without bound).

Then add 1: \( -2f(x)+1<0 + 1=1 \). Also, as \( f(x) ightarrow0^+ \), \( -2f(x)+1 ightarrow - 2\times0+1 = 1 \), and as \( f(x) ightarrow\infty \), \( -2f(x)+1 ightarrow-\infty \). So the range of \( g(x) \) is all real numbers less than 1, i.e., \( (-\infty, 1) \), which is option D. Wait, but let's check with an example. When \( x = 0 \), \( f(0)=10^0 = 1 \), \( g(0)=-2\times1 + 1=-1 \). When \( x = 1 \), \( f(1)=10 \), \( g(1)=-2\times10 + 1=-19 \). When \( x=-1 \), \( f(-1)=\frac{1}{10} \), \( g(-1)=-2\times\frac{1}{10}+1=-\frac{1}{5}+1=\frac{4}{5}=0.8 \). So as \( x \) decreases (goes to \( -\infty \)), \( f(x)=10^x \) approaches 0, so \( g(x) \) approaches \( -2\times0 + 1 = 1 \). So the values of \( g(x) \) are all real numbers less than 1. So the range is \( (-\infty, 1) \), which is option D. Wait, but earlier when I thought I made a mistake, maybe I confused the steps. Let's confirm the inequality again.

Given \( f(x)>0 \)

Multiply both sides by - 2: \( -2f(x)<0 \) (because \( -2\times positive< - 2\times0 = 0 \))

Add 1 to both sides: \( -2f(x)+1<0 + 1\Rightarrow g(x)<1 \)

And since \( f(x) \) can be made arbitrarily large (so \( -2f(x) \) can be made arbitrarily small, i.e., go to \( -\infty \)), \( g(x) \) can take any value less than 1. So the range is \( (-\infty, 1) \), which is option D.

Wait, but the options:

A. \( (-2, \infty) \)

B. \( (-\infty, 2) \)

C. \( (0, \infty) \)

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

Yes, so the correct answer is D.

Answer:

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