Question

Given is the graph of the function h(x)=log1/2 x
. Use the graph to find h(2)+h−1(−2)
.

h(2)+h−1(−2)

Answer 1

To solve for \( h(2) + h^{-1}(-2) \) given the function \( h(x) = \log_{\frac{1}{2}} x \), we can approach this step-by-step.

### Step 1: Calculate \( h(2) \)

The function is given as \( h(x) = \log_{\frac{1}{2}} x \). We can use the change of base formula to compute \( h(2) \):

\[
h(2) = \log_{\frac{1}{2}} 2 = \frac{\log 2}{\log \frac{1}{2}} = \frac{\log 2}{-1} = -\log 2
\]

### Step 2: Find the inverse \( h^{-1}(-2) \)

To find \( h^{-1}(-2) \), we first need to determine \( y \) such that \( h(y) = -2 \):

\[
h(y) = \log_{\frac{1}{2}} y = -2
\]

We can convert this logarithmic equation to its exponential form. The equation \( \log_{\frac{1}{2}} y = -2 \) can be rewritten as:

\[
y = \left( \frac{1}{2} \right)^{-2} = 2^2 = 4
\]

Thus, \( h^{-1}(-2) = 4 \).

### Step 3: Combine the results

Now we can find \( h(2) + h^{-1}(-2) \):

\[
h(2) + h^{-1}(-2) = (-\log 2) + 4 = 4 - \log 2
\]

### Final Answer

Thus, the final result is:

\[
h(2) + h^{-1}(-2) = 4 - \log 2
\]