Let \(f\) be the function that assigns to a temperature in degrees Celsius its equivalent in degrees Fahrenheit.
Our function, \(f\), described by the equation $$ y = \frac{9}{5}x + 32 $$ takes a temperature in degrees Celsius and gives its equivalent in degrees Fahrenheit. If we wish to have a function that takes a temperature in degrees Fahrenheit and gives its equivalent in degrees Celsius we switch the roles of \(x\) and \(y\) in the equation above, and solve for \(x\) in terms of \(y\). This means subtracting \(32\) from both sides of the equation and then multiplying both sides of the new equation by \(\frac{5}{9}\), giving: $$ x = \frac{5}{9}\left(y - 32\right). $$ We are used to \(y\) being the dependent variable and \(x\) being the independent variable so we can rewrite this, using function notation, as $$ g(x) = \frac{5}{9}\left(x-32\right) $$ and this function converts the temperature in degrees Fahrenheit to the corresponding temperature in degrees Celsius.
Commentary
Temperature conversions provide a rich source of linear functions which are encountered not only in science but also in our every day lives when we travel abroad. The first part of this task provides an opportunity to construct a linear function given two input-output pairs. The second part investigates the inverse of a linear function while the third part requires reasoning about quantities and/or solving a linear equation.
In part (c), students could also argue try to give an intuitive argument for the existence of such a point, reasoning via the linear relationship between degrees Fahrenheit (F) and degrees Celsius (C). Namely, we can first easily check that \(F\lt C\) when, say, \(C=-100\), and that \(F\gt C\) when, for example, \(C=0\). Since there is a linear relationship between \(F\) and \(C\), there must then be some value of \(C\) between \(C=-100\) and \(C=0\) where in fact \(F=C\). This is a subtle but important argument, one that eventually leads to more advanced topics like continuity and the Intermediate Value Theorem.