Question

consider the three functions.
$f(x) = -\frac{1}{11} left( \frac{9}{7}
ight)^x$
$g(x) = \frac{1}{11} left( \frac{9}{7}
ight)^{-x}$
$h(x) = -\frac{1}{11} left( \frac{9}{7}
ight)^{-x}$
which statement is true?

• the range of $h(x)$ is $y > 0$.
• the domains of $f(x)$ and $g(x)$ are different from the domain of $h(x)$.
• the domain of $g(x)$ is $y > 0$.
• the ranges of $f(x)$ and $h(x)$ are different from the range of $g(x)$.

Explanation:

Step1: Analyze Domain of Exponential Functions

Exponential functions of the form \( a^x \) (or \( a^{-x}=( \frac{1}{a})^x \)) have a domain of all real numbers (\( \mathbb{R} \)) because we can substitute any real number for \( x \). So, \( f(x)=-\frac{8}{11}(\frac{9}{5})^x \), \( g(x)=\frac{8}{11}(\frac{9}{5})^{-x} \), and \( h(x)=-\frac{8}{11}(\frac{9}{5})^{-x} \) all have domain \( \mathbb{R} \) (all real numbers). This means the statement "The domains of \( f(x) \) and \( g(x) \) are different from the domain of \( h(x) \)" is false, and the statement "The domain of \( g(x) \) is \( y > 0 \)" is also false (since domain is about \( x \)-values, not \( y \)-values, and \( g(x) \)'s domain is all real numbers).

Step2: Analyze Range of Exponential Functions

For an exponential function \( a^x \) where \( a>0,a
eq1 \), the range is \( y > 0 \) (since any positive number raised to a real power is positive). Let's analyze each function:

• For \( f(x)=-\frac{8}{11}(\frac{9}{5})^x \): \( (\frac{9}{5})^x>0 \) for all real \( x \), so \( -\frac{8}{11}(\frac{9}{5})^x<0 \) (multiplying a positive number by \( -\frac{8}{11} \) (negative) gives a negative number). So the range of \( f(x) \) is \( y < 0 \).
• For \( g(x)=\frac{8}{11}(\frac{9}{5})^{-x}=\frac{8}{11}(\frac{5}{9})^x \): \( (\frac{5}{9})^x>0 \) for all real \( x \), so \( \frac{8}{11}(\frac{5}{9})^x>0 \) (multiplying a positive number by \( \frac{8}{11} \) (positive) gives a positive number). So the range of \( g(x) \) is \( y > 0 \).
• For \( h(x)=-\frac{8}{11}(\frac{9}{5})^{-x}=-\frac{8}{11}(\frac{5}{9})^x \): \( (\frac{5}{9})^x>0 \) for all real \( x \), so \( -\frac{8}{11}(\frac{5}{9})^x<0 \) (multiplying a positive number by \( -\frac{8}{11} \) (negative) gives a negative number). So the range of \( h(x) \) is \( y < 0 \), which means the statement "The range of \( h(x) \) is \( y > 0 \)" is false.

Step3: Compare Ranges

Now, compare the ranges:

• Range of \( f(x) \): \( y < 0 \)
• Range of \( g(x) \): \( y > 0 \)
• Range of \( h(x) \): \( y < 0 \)

So the ranges of \( f(x) \) and \( h(x) \) (both \( y < 0 \)) are different from the range of \( g(x) \) ( \( y > 0 \) ).

Answer:

The ranges of \( f(x) \) and \( h(x) \) are different from the range of \( g(x) \).