Answer :
Certainly! Let's evaluate the function [tex]\( h(x) = \left(\frac{1}{10}\right)^x \)[/tex] for each given [tex]\( x \)[/tex] value step by step.
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[ h(-2) = \left(\frac{1}{10}\right)^{-2} \][/tex]
When a base with a negative exponent is found, we take the reciprocal of the base and change the exponent to positive:
[tex]\[ \left(\frac{1}{10}\right)^{-2} = 10^2 = 100 \][/tex]
Therefore,
[tex]\[ h(-2) = 100 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ h(-1) = \left(\frac{1}{10}\right)^{-1} \][/tex]
Similarly, taking the reciprocal of the base and changing the exponent to positive:
[tex]\[ \left(\frac{1}{10}\right)^{-1} = 10^1 = 10 \][/tex]
Therefore,
[tex]\[ h(-1) = 10 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \left(\frac{1}{10}\right)^0 \][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[ \left(\frac{1}{10}\right)^0 = 1 \][/tex]
Therefore,
[tex]\[ h(0) = 1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = \left(\frac{1}{10}\right)^1 \][/tex]
Any number raised to the power of 1 is the number itself:
[tex]\[ \left(\frac{1}{10}\right)^1 = \frac{1}{10} = 0.1 \][/tex]
Therefore,
[tex]\[ h(1) = 0.1 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ h(2) = \left(\frac{1}{10}\right)^2 \][/tex]
Squaring the base 1/10:
[tex]\[ \left(\frac{1}{10}\right)^2 = \frac{1}{100} = 0.01 \][/tex]
Therefore,
[tex]\[ h(2) = 0.01 \][/tex]
Summarizing the results, we have:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $h(x)$ \\ \hline -2 & 100 \\ \hline -1 & 10 \\ \hline 0 & 1 \\ \hline 1 & 0.1 \\ \hline 2 & 0.01 \\ \hline \end{tabular} \][/tex]
1. For [tex]\( x = -2 \)[/tex]:
[tex]\[ h(-2) = \left(\frac{1}{10}\right)^{-2} \][/tex]
When a base with a negative exponent is found, we take the reciprocal of the base and change the exponent to positive:
[tex]\[ \left(\frac{1}{10}\right)^{-2} = 10^2 = 100 \][/tex]
Therefore,
[tex]\[ h(-2) = 100 \][/tex]
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ h(-1) = \left(\frac{1}{10}\right)^{-1} \][/tex]
Similarly, taking the reciprocal of the base and changing the exponent to positive:
[tex]\[ \left(\frac{1}{10}\right)^{-1} = 10^1 = 10 \][/tex]
Therefore,
[tex]\[ h(-1) = 10 \][/tex]
3. For [tex]\( x = 0 \)[/tex]:
[tex]\[ h(0) = \left(\frac{1}{10}\right)^0 \][/tex]
Any non-zero number raised to the power of 0 is 1:
[tex]\[ \left(\frac{1}{10}\right)^0 = 1 \][/tex]
Therefore,
[tex]\[ h(0) = 1 \][/tex]
4. For [tex]\( x = 1 \)[/tex]:
[tex]\[ h(1) = \left(\frac{1}{10}\right)^1 \][/tex]
Any number raised to the power of 1 is the number itself:
[tex]\[ \left(\frac{1}{10}\right)^1 = \frac{1}{10} = 0.1 \][/tex]
Therefore,
[tex]\[ h(1) = 0.1 \][/tex]
5. For [tex]\( x = 2 \)[/tex]:
[tex]\[ h(2) = \left(\frac{1}{10}\right)^2 \][/tex]
Squaring the base 1/10:
[tex]\[ \left(\frac{1}{10}\right)^2 = \frac{1}{100} = 0.01 \][/tex]
Therefore,
[tex]\[ h(2) = 0.01 \][/tex]
Summarizing the results, we have:
[tex]\[ \begin{tabular}{|c|c|} \hline $x$ & $h(x)$ \\ \hline -2 & 100 \\ \hline -1 & 10 \\ \hline 0 & 1 \\ \hline 1 & 0.1 \\ \hline 2 & 0.01 \\ \hline \end{tabular} \][/tex]