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The formula $ C = \frac{5}{9} (F - 32) $ , where $ F \ge -459.67 $, expresses the Celsius temperature $ C $ as a function of the Fahrenheit temperature $ F $. Find a formula for the inverse function and interpret it. What is the domain of the inverse function?

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We solve $C=\frac{5}{9}(F-32)$ for $F: \frac{9}{5} C=F-32 \Rightarrow F=\frac{9}{5} C+32 .$ This gives us a formula for the inverse function, that is, the Fahrenheit temperature $F$ as a function of the Celsius temperature $C . F \geq-459.67 \Rightarrow \frac{9}{5} C+32 \geq-459.67 \Rightarrow$ $\frac{9}{5} C \geq-491.67 \Rightarrow C \geq-273.15,$ the domain of the inverse function.

02:49

Jeffrey Payo

Calculus 1 / AB

Calculus 2 / BC

Calculus 3

Chapter 1

Functions and Models

Section 5

Inverse Functions and Logarithms

Functions

Integration Techniques

Partial Derivatives

Functions of Several Variables

Tuv D.

August 25, 2020

The Domain is not C>=-626.06, it's C>=-273.15

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all right, so here's a formula that relates apparent high temperature and the Celsius temperature, and we're going to find the inverse of it, which basically means, in this context, isolate F instead of having see isolated. And so I'm not going to switch the variables like normally, we would switch X and Y. But if I switch X if I switch C and F, then they're no longer going to mean what we want them to mean. We don't want to see to stand for fair and height and f to stand for Celsius, so I'm not going to switch the variables in the equation. I'm just going to isolate the other one, so I'll start by multiplying both sides by 9/5. So 9/5 C equals F minus 32 and then add 32 to both sides. So ninth of C plus 32 equals F. So this would be what we would call the inverse then. Technically, we didn't do any switching our inner changing of variables, but we isolated the other variable so we can interpret this as a fair in high temperature with respect to the Celsius temperature or as a function of the Celsius temperature. Now we're interested in knowing the domain of this. Well, the domain of this function will be whatever the range of the previous function Waas. So what was the range of the previous function? If the domain was f is greater than or equal to negative 4 59.67 and we substitute that into the equation and sell for see, we end up with negative 6 26.6 Soc is greater than or equal to negative 6 26.6 That was the range of the original function, So it will be the domain of the inverse function.

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