To solve for \( h(2) + h^{-1}(-2) \), we first need to find \( h(2) \) and \( h^{-1}(-2) \).
Step 1: Calculate \( h(2) \)
Given the function \( h(x) = \log_{12}(x) \), we can calculate \( h(2) \):
\[ h(2) = \log_{12}(2) \]
This can be simplified using the change of base formula:
\[ h(2) = \frac{\log_{10}(2)}{\log_{10}(12)} \]
Or if using natural logarithms:
\[ h(2) = \frac{\ln(2)}{\ln(12)} \]
Step 2: Find \( h^{-1}(-2) \)
The value \( h^{-1}(-2) \) represents the x value for which \( h(x) = -2 \):
\[ h(x) = -2 \Rightarrow \log_{12}(x) = -2 \]
To solve for \( x \), we rewrite this equation in its exponential form:
\[ x = 12^{-2} = \frac{1}{12^2} = \frac{1}{144} \]
So, \( h^{-1}(-2) = \frac{1}{144} \).
Step 3: Combine the results
Now, we need to calculate \( h(2) + h^{-1}(-2) \):
\[ h(2) + h^{-1}(-2) = \log_{12}(2) + \frac{1}{144} \]
If you have numerical values for \( \log_{10}(2) \) and \( \log_{10}(12) \), you can compute \( h(2) \) numerically. However, \( \log_{12}(2) \) alone cannot be simplified further without specific logarithm values.
Final Calculation
\[ \text{Final Result} = \log_{12}(2) + \frac{1}{144} \]
If you need numeric approximations, you could use values like:
\(\log_{10}(2) \approx 0.3010\) and \(\log_{10}(12) \approx 1.0792\):
\[ h(2) \approx \frac{0.3010}{1.0792} \approx 0.278 \]
Thus,
\[ h(2) + h^{-1}(-2) \approx 0.278 + \frac{1}{144} \approx 0.278 + 0.0069 = 0.2849 \text{ (approximately)} \]
Thus, the answer expresses as the combination of both results.