Exam1reviewAnswers - Math 112 1 Find the domain of the...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 112 Exam 1 Review Fall 2010 1. Find the domain of the functions (a) g ( x ) = 3 - x ( x +1) 2 Answer: ( -∞ , - 1) ( - 1 , - 3] (b) log( x - 3 - 4 /x ) Answer: ( - 1 , 0) (4 , ) 2. The graph of y = F ( x ) is given. Sketch the graph of the following functions: y = F ( x ) 6 - x y Q Q Q Q Q 1 1 (a) y = F ( x + 2) - 3 Answer: 6 - x y Q Q Q Q Q 1 1
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(b) y = F (1 - x ) Answer: 6 - x y S S S S S S P P P P P 1 1 3. Graph f ( x ) = log( x - 3) + 2. State the x - intercept and the asymptote. Answers: x - intercept: x = 3 . 01 Asymptote: x = 3 4. Evaluate exactly: (a) log 2 3 64 Answer: 2 (b) log 3 16 Answer: 4 log 2 / log 3 (c) 5 log 5 125 Answer: 125 (d) ln e 2 - ln e 4 Answer: - 2 5. Sketch the function f ( x ) = 2 x +3 if - 4 x < 0 - 2 x + 8 if x 0 Then use your graph to answer the following questions: (a) is f one-to-one? Answer: No (b) f ( - 2) = Answer: 2 (c) f (0) = Answer: 8 (d) f (2) = Answer: 4 (e) f ( - 5) = Answer: Not defined
Image of page 2
6. Solve each equation exactly: (a) log 3 ( x + 2) - log 3 ( x - 1) = 3 Answer: 29/26 (b) log 5 (2 x - 1) + log 5 ( x + 4) = log 5 x Answer: ( 17 - 3) / 2 (c) 3 e 2 x - 1 = 12 Answer: (ln 4 + 1) / 2 (d) 2 x +2 = 5 3 x Answer: (2 ln 2) / (3 ln 5 - ln 2) 7. Find the inverse of the functions: (a) f ( x ) =
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern