question 1 of 13 (1 point) | question attempt: 1 of unlimited\n$g(x)=\\left(\\frac{1}{9}\\right)^x$\nfind…

Explanation:

Step1: Correct the function form

Assume the function is $g(x) = \left(\frac{1}{9}\right)^x$ (standard exponential form).

Step2: Calculate $g(-2)$

Use negative exponent rule: $a^{-n}=\frac{1}{a^n}$ $g(-2) = \left(\frac{1}{9}\right)^{-2} = 9^2 = 81$

Step3: Calculate $g(-1)$

Apply negative exponent rule $g(-1) = \left(\frac{1}{9}\right)^{-1} = 9^1 = 9$

Step4: Calculate $g(0)$

Use zero exponent rule: $a^0=1$ $g(0) = \left(\frac{1}{9}\right)^0 = 1$

Step5: Calculate $g(1)$

Evaluate positive exponent $g(1) = \left(\frac{1}{9}\right)^1 = \frac{1}{9}$

Step6: Calculate $g(2)$

Evaluate positive exponent $g(2) = \left(\frac{1}{9}\right)^2 = \frac{1}{81}$

Answer:

$x$	$g(x)$
$-2$	$81$
$-1$	$9$
$0$	$1$
$1$	$\frac{1}{9}$
$2$	$\frac{1}{81}$