Stitz-Zeager_College_Algebra_e-book

Example 654 a 40 f roast is cooked in a 350 f oven

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Unformatted text preview: call from Example 6.1.2 that the temperature of coffee T (in degrees Fahrenheit) t minutes after it is served can be modeled by T (t) = 70 + 90e−0.1t . When will the coffee be warmer than 100◦ F? Solution. We need to find when T (t) > 100, or in other words, we need to solve the inequality 70 + 90e−0.1t > 100. Getting 0 on one side of the inequality, we have 90e−0.1t − 30 > 0, and we set r(t) = 90e−0.1t − 30. The domain of r is artificially restricted due to the context of the problem to [0, ∞), so we proceed to find the zeros of r. Solving 90e−0.1t − 30 = 0 results in e−0.1t = 1 so that t = −10 ln 1 which, after a quick application of the Power Rule leaves us with 3 3 t = 10 ln(3). If we wish to avoid using the calculator to choose test values, we note that since 1 < 3, 0 = ln(1) < ln(3) so that 10 ln(3) > 0. So we choose t = 0 as a test value in [0, 10 ln(3)). Since 3 < 4, 10 ln(3) < 10 ln(4), so the latter is our choice of a test value for the i...
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