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The given problem shows the relationship between Fahrenheit \( F \) and Celsius \( C \) defined by the equation:

\[
F = g(C) = \frac{9}{5}C + 32
\]

### (a) Find \( g^{-1} \)

To find the inverse function \( g^{-1}(F) \), we need to solve for \( C \) in terms of \( F \) from the given equation:

\[
F = \frac{9}{5}C + 32
\]

Steps:
1. Subtract 32 from both sides:
   \[
   F - 32 = \frac{9}{5}C
   \]
   
2. Multiply both sides by \( \frac{5}{9} \):
   \[
   C = \frac{5}{9}(F - 32)
   \]

Thus, the inverse function \( g^{-1}(F) \) is:

\[
g^{-1}(F) = \frac{5}{9}(F - 32)
\]

### (b) Find \( g^{-1}(113) \)

Now we need to calculate \( g^{-1}(113) \) using the inverse function:

\[
g^{-1}(113) = \frac{5}{9}(113 - 32) = \frac{5}{9}(81) = \frac{405}{9} = 45
\]

Therefore:

\[
g^{-1}(113) = 45
\]

### Interpretation:
- For part (a), \( g^{-1}(F) \) represents the Celsius temperature that corresponds to a Fahrenheit temperature.
- For part (b), \( g^{-1}(113) \) represents the Celsius temperature corresponding to \( 113^\circ F \), which is \( 45^\circ C \).

Would you like more details or have any other questions?

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Here are 5 related questions to expand on this:

1. How do you calculate the Fahrenheit temperature from a given Celsius temperature?
2. What is the significance of the slope \( \frac{9}{5} \) in the Fahrenheit-Celsius conversion equation?
3. How do you derive inverse functions from linear equations in general?
4. What is the physical meaning of the intercept 32 in the equation \( F = \frac{9}{5}C + 32 \)?
5. Can you generalize this approach to derive inverse functions for other temperature scales like Kelvin?

**Tip:** Always check the units when converting between temperature scales to ensure you're interpreting the results correctly.

The relationship between Fahrenheit (F) and Celsius (C) scales is given by F = g(C) = (9/5)C + 32. (a) Find g^{-1}. (b) Find g^{-1}(113). What do the inverse functions represent?

Learn how to find the inverse function g^{-1} for converting Fahrenheit to Celsius using the equation F = (9/5)C + 32. This problem demonstrates the calculation of g^{-1}(F) and applies it to g^{-1}(113) to find the Celsius equivalent. Suitable for students in Grades 9-11.

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