Question

Given is the graph of the function h(x)=log12x . Use the graph to find h(2)+h−1(−2) .

Answer 1

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.