Midterm #1B. x 8 < < x 8 < 11 3 < x < x > x < 5 or 3 2x > 5 2x < 8 2x > 2

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1 Mat 30 College Algebra Februar 2, 2016 Midterm #1B Name: Answer Ke David Arnold Instructions. ( points) For eac o te ollowing questions, select te best answer and darken te corresponding circle on our scantron. ( pts ) 1. Wic o te ollowing best describes te solution o 8 < 11? (a) ( 19, 3) (b) (, 3) (19, ) (c) (, 19) (d) ( 3, 19) (e) (, 19) (3, ) Solution: Note tat < a is equivalent to a < < a, so: Sketc te solution on a number line. Using interval notation, te solution is ( 3, 19). 8 < < 8 < 11 3 < < ( pts ) 2. Wic o te ollowing best describes te solution o 3 2 >? (a) (, 4) (1, ) (b) ( 1, 4) (c) (, 1) (d) (, 1) (4, ) (e) ( 4, 1) Solution: Note tat > a is equivalent to < a or > a, so: becomes 3 2 > 3 2 < or 3 2 > 2 < 8 2 > 2 > 4 Sketc te solution on a number line. 1 4 Using interval notation, te solution is (, 1) (4, ). < 1 ( pts ) 3. Given () = 19 2, evaluate (3 2). (a) (b) (c) (d) (e) Solution: Given () = 19 2, ten: (3 2) = 19 (3 2) 2 = 19 ( ) =

2 Mat 30 College Algebra/Midterm #1B Page 2 o 13 Name: Answer Ke ( pts ) 4. Given () = and g() = 9 2, evaluate ( + g)(2). (a) 12 (b) (c) 20 (d) (e) 3 Solution: Given () = and g() = 9 2, ten: ( + g)(2) = (2) + g(2) = [2(2) + 3] [9 (2) 2 ] = 7 + = 12 ( pts ). Use interval notation to describe te domain o () = (a) (, 1] (b) [1, ) (c) (, 1) (d) (1, ) (e) (, 2) ( 2, 1) Solution: We cannot take te square root o a negative number, so te epression 1 must be greater tan or equal to zero. However, because 1 is in te denominator, 1 must be strictl greater tan zero. Sading on a number line: 1 > 0 > 1 < 1 Using interval notation, te domain is (, 1). 1 ( pts ) 6. Given te grap o, identi te solution o () 0. (a) (, 3] [1, ) (b) ( 1, 3) (c) [ 1, 3] (d) (, 3] (e) (, 1] [3, ) Solution: Draw dased vertical line troug te -intercepts.

3 Mat 30 College Algebra/Midterm #1B Page 3 o 13 Name: Answer Ke 1 3 Net, note tat te grap o is below or on te -ais outside te dased lines, as saded in te above igure. In interval notation, te solution is ( 1] [3, ). ( pts ) 7. Given te grap o, identi te range o. (a) (, 2] (b) (, ) (c) [ 2, 1] (d) [0, ) (e) [ 2, ) Solution: I we project all te points on te grap o onto te -ais, te result is sown in te ollowing igure. Using interval notation, te range is [ 2, ).

4 Mat 30 College Algebra/Midterm #1B Page 4 o 13 Name: Answer Ke ( pts ) 8. Given te grap o, evaluate ( 4). (a) 2 (d) Undeined (b) 0 (c) 2 (e) 4 Solution: Note tat te grap o intercepts te -ais at ( 4, 0), or ( 4, ( 4)). ( 4, 0) Hence, ( 4) = 0. ( pts ) 9. Given evaluate ( 3). (a) 9 (d) 2, i < 0 () =, i 0 < , i 3 (b) 9 (c) 3 (e) Solution: Because 3 < 0, we use te piece () = 2 to ind te answer: ( 3) = ( 3) 2 = 9

5 Mat 30 College Algebra/Midterm #1B Page o 13 Name: Answer Ke ( pts ) 10. Consider te grap o. (a) is an even unction (b) is an odd unction (c) is neiter an even nor an odd unction Solution: Consider te grap o. Note tat te grap is smmetric wit respect to te origin. Hence, is an odd unction. ( pts ) 11. Wic o te ollowing best describes te set o all -values were te grap o is increasing?

6 Mat 30 College Algebra/Midterm #1B Page 6 o 13 Name: Answer Ke (a) (, 2) (2, ) (b) ( 1, 1) (c) (, 1) (1, ) (d) ( 2, 2) (e) (1, ) Solution: Te grap o is increasing as indicated in te ollowing image. However, we re asked or wat -values is te grap increasing. We sade tese -values on te -ais. 1 1 Hence, te grap o is increasing on te interval (, 1) (1, ). Instructions. (70 points) Answer eac o te ollowing questions on our own paper. Wen inised, arrange tese solutions in order, ten place our eam on top and staple te entire package. (10 pts ea. ) 1. Solve eac o te ollowing equations or. (a) 2/ 2 1/ = 3 Solution: Te equation 2/ 2 1/ = 3 is a disguised quadratic. Let u = 1/. Ten u 2 = 2/. Make tese substitutions, ten te equation becomes: u 2 2u = 3 Tis equation is nonlinear, so start b making one side zero, ten actoring. u 2 2u 3 = 0 (u 3)(u + 1) = 0

7 Mat 30 College Algebra/Midterm #1B Page 7 o 13 Name: Answer Ke Hence, te solutions are: u = 3 or u = 1 Now we replace u wit 1/ and solve or. 1/ = 3 = 243 1/ = 1 = 1 (b) = 1 Solution: Te equation = a, were a > 0 is equivalent to = a or = a. Hence becomes: = = 1 or = 1 Bot equations are nonlinear, so move all terms to one side o te equations = = 0 ( + 7)( 1) = 0 = 6 ± = 7, 1 = 6 ± = 3 ± 3 2 (10 pts ) 2. Given te unction () = , simpli te epression as muc as possible. Solution: ( + ) () ( + ) () Tis answer is valid, providing 0. = [3( + )2 + ( + ) 2] [ ] = [3( ) + ( + ) 2] [ ] = = ( = =

8 Mat 30 College Algebra/Midterm #1B Page 8 o 13 Name: Answer Ke (10 pts ) 3. Consider te unction g() = Use a Matematica to sketc te grap o g, adjusting te domain so tat all -intercepts are visible. Use Matematica s FindRoot command to identi te coordinates o te -intercepts. On grap paper, perorm eac o te ollowing tasks. (a) Cop te Matematica grap onto grap paper. Label and scale eac ais. Solution: Enter te ollowing code. Clear[g, ] g[_] = 0.4 ^ ; Plot[g[], {, -7, 1}, AesLabel -> {"", ""}, GridLines -> Automatic] Tis produces te ollowing image wic sould be drawn on grap paper (b) Label eac -intercept wit its coordinates, ten draw dased vertical lines troug eac -intercept. Solution: Te ollowing commands provide us wit te -intercepts. In[12]:= FindRoot[g[], {, -3}] Out[12]= { -> } In[126]:= FindRoot[g[], {, 13}] Out[126]= { -> } Te ollowing code adds te vertical dased line and labels te -intercepts. Plot[g[], {, -7, 1}, AesLabel -> {"", ""}, GridLines -> Automatic, Epilog -> { Red, Tet[" ", { , -}], Tet[" ", { , -}], Dased, IniniteLine[{{ , 0}, { , 1}}], IniniteLine[{{ , 0}, { , 1}}] }] Wic produces tis image.

9 Mat 30 College Algebra/Midterm #1B Page 9 o 13 Name: Answer Ke (c) On te -ais, sade te solution o g() 0. Solution: Te question asks us to ind te solution o g() 0. Tis means we must located were te grap o g lies on and below te -ais. Te answer is on and outside te vertical dased lines. We can sade tis region wit te ollowing code. Plot[g[], {, -7, 1}, AesLabel -> {"", ""}, GridLines -> Automatic, Epilog -> { Red, Tet[" ", { , -}], Tet[" ", { , -}], {Dased, IniniteLine[{{ , 0}, { , 1}}], IniniteLine[{{ , 0}, { , 1}}]}, Tickness[0.01], Arrow[{{ , 0}, {-7, 0}}], Arrow[{{ , 0}, {1, 0}}], PointSize[Large], Point[{{ , 0}, { , 0}}] }] Wic produces te ollowing image (d) Use set-builder and interval notation to describe te solution o g() 0. Solution: Using set-builder notation, we describe te sading on te -ais b: { : } Using interval notation, we describe te sading on te -ais b: [ , ]

10 Mat 30 College Algebra/Midterm #1B Page 10 o 13 Name: Answer Ke (10 pts ) 4. Prove tat te unction () = is even. Note: To receive an credit or tis proo, ou must present our proo as demonstrated in class. Solution: To sow tat is even, we need to sow tat ( ) = () or all. Proo: I () = , ten: ( ) = 3 2( ) 2 + = = () Tereore, is even. (10 pts ). Cop te given grap o onto grap paper, using a blue pencil to draw te grap o. Label and scale our aes appropriatel Draw a sequence o graps, one step at a time, eac labeled wit te equation o te grap at tat step, used to produce te inal grap o = ( ) + 2. Solution: Start wit te original grap:

11 Mat 30 College Algebra/Midterm #1B Page 11 o 13 Name: Answer Ke Now sketc = ( ), wic is a relection o te original unction across te -ais = ( ) Net, add 2 to obtain = ( ) + 2. Tis will move te grap two units upward.

12 Mat 30 College Algebra/Midterm #1B Page 12 o 13 Name: Answer Ke 3 = ( ) Hence, te inal grap as te equation = ( ) + 2. (10 pts ) 6. Consider te unction () = (a) Set up a number line. Place te critical value or eac epression inside te absolute value bars on our number line. List te epressions inside te absolute bars below te number line at te let end o te number line. Use plus and minus signs to indicate te sign o eac epression in te intervals determined b te critical values. Above te number line, list eac epression surrounded b te absolute value bars at te let end o te number line. Sow te results o eac epression wen te absolute value bars are removed above te number line in eac region determined b te critical values. Solution: Set up a number line and use te point-testing metod to determine te sign o eac epression. Use te result to remove te absolute value bars on eac interval determined b te critical values o te epressions inside te absolute value bars (b) Create a piecewise deinition o te unction. Solution: Net, sum eac o te epressions on eac interval determined b te critical values , i < 1 () = 6, i 1 < 7 2 8, i 7 (c) Witout te aid o a calculator, sketc te grap o on a seet o grap paper. Label and scale eac ais and label te grap wit its equation. Solution: Te irst piece is = 2 + 8, a ra starting at = 1 and moving to te let. Te second piece = 6 is a orizontal line segment connecting te points (1, 6) and (7, 6). Te last piece = 2 8 is a ra starting at = 7 and moving to te rigt.

13 Mat 30 College Algebra/Midterm #1B Page 13 o 13 Name: Answer Ke