MATH 151 Midterm: MATH 151 TAMU Y2010 2010a Exam 3b Solutions
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Exam iii version b solutions: a f (x) = 3(x 3)2 = 0 when x = 3. Test- ing the endpoints and the critical value in the original function yield f (1) = 8, f (3) = 0, f (4) = 1, so the absolute maximum is 1 when x = 4: c y = log27 9 is equivalent to 27y = 9, or (alterna- 33y = 32, so 3y = 2 and y = tively, use the change-of-base formula). 3: d using properties of logarithms, we have log10((5 x)(2 x)) = 1, which is equiva- lent to 10 7x + x2 = 101, or x2 7x = 0. A quick check shows that x = 7 is not in the domain of the original equation, so the only solution is x = 0: c y = sin 1(cid:16) x. From the reference triangle below, we see that cos y = 1 + x6 , so f (1) =
