Chapter 6 Rational Exponents and Radical Functions - 6.4 Use Inverse Functions - 6.4 Exercises - Problem Solving - Page 444: 48b

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Substitute $C=5$: $C=\frac{5}{9}(F-32)\\5=\frac{9}{5}(F-32)\\9=F-32\\F=41$ The Fahrenheit temperature at the start of the race is $41$ degrees. At the end of the race, the temperature was $-10$ degrees. Substitute $C=-10$: $C=\frac{5}{9}(F-32)\\-10=\frac{9}{5}(F-32)\\-18=F-32\\F=14$ The Fahrenheit temperature at the end of the race is $14$ degrees.