Name: Ann Khadaran
Student I.D. # 0406235
Telephone # (908) 654-6795
Email address:
College and Semester: TESC, September 2008
Course Code: 2008SEP MAT-121-GS008GS
Course Name: College Algebra
Written Assignment # 6
Determine the intervals of the domain over which each function is continuous.
Graph referred to are in the attached document
2.5 #8Which one is the graph of ? On what interval is it increasing?
Ans: Graph GIt is increasing on the interval x > 0
2.5 #10Which one is not the graph of a function? What is its equation?
Ans: Graph CIts equation is
2.5 #12Which one is the graph of ? What is the value of y when x = 1.5?
Ans: Graph BThe value of y is 1
2.5 #16Which graphs of functions decrease over part of the domain and increase over the rest of the domain? On what intervals do they increase? Decrease?
Ans: Graphs E and GThey both decrease on the interval x<0 and increase
on the interval x>0.
2.5 # 18For each piecewise-defined function, find (a) f(-5), (b) f(-1), (c) f(0), and (d) f(3).
Ans:(a)
(b)
(c)
(d)
2.5 #40Match each piecewise-defined function with its calculator graph. (All graphs are shown in dot mode.)
Ans: Graph B No, Graph A
2.5 #42Match each piecewise-defined function with its calculator graph. (All graphs are shown in dot mode.)
Ans: Graph C
2.6 #22Without graphing, determine whether each equation has a graph that is symmetric with respect to the x-axis, the y-axis, the origin, or none of these.
Ans: Symmetric with respect to y-axis and the origin
This one I find a bit of a puzzle. If the -5 were a +5, then the answer would be symmetric with respect to the y-axis (since plugging in –x gives the same equation), the x-axis (since plugging in –y gives the same equation), and the origin (since plugging in both –x and –y gives the same equation). So all three symmetries would be in the graph. However, the -5 poses a problem. Rearranging the equation a bit, we get x2 y2 = -5. If you take the product of the squares of two numbers, you will not get a negative answer! Therefore, this equation has no real solutions. Therefore, it has no graph on a real coordinate system. Since there is no graph, there are no symmetries of that graph.
2.6 #26Without graphing, determine whether each equation has a graph that is symmetric with respect to the x-axis, the y-axis, the origin, or none of these.
y = x + 12
y = x + 12
y = -x + 12
y = x + 12
-y = x + 12
y = x + 12
-y = -x + 12
Ans: None of these
2.6 # 42Graph each function.
Ans: Please see attached for graph
I didn’t see an attached graph, but the graph is below for you to compare.
Let and . Find each of the following. This pertains to the next 4 questions
2.7 #2
Ans:
(f – g)(-5) actually means f(-5) – g(-5) = 135 – (-26) = 161
f(-5) = 5 (-5)2 - 2(-5) = 5(25) + 10 = 125 + 10 = 135
g(-5) = 6(-5) + 4 = -30 + 4 = -26
2.7 #4
Ans:
(fg)(-3) = f(-3) * g(-3) = 51 * (-14) = -714
f(-3) = 5 (-3)2 – 2(-3) = 5 (9) + 6 = 45 + 6 = 51
g(-3) = 6(-3) + 4 = -18 + 4 = -14
2.7 #6
Ans:
(f/g)(4) = f(4) / g(4) = 72 / 28 = 18 / 7
f(4) = 5(4)2 – 2(4) = 5 (16) – 8 = 80 – 8 = 72
g(4) = 6(4) + 4 = 24 + 4 = 28
2.7 #8
Ans:
(f+g)(2k) = f(2k) + g(2k) = 20k2 – 4k + 12k + 4 = 20k2 + 8k + 4
f(2k) = 5 (2k)2 – 2(2k) = 5(4k2) – 4k = 20k2 - 4k
g(2k) = 6(2k) + 4 = 12k + 4
2.7 #36For each of the functions defined as follows, find (a), (b), and (c) .
The function definitions are missing, but I will do an example.
Let’s take f(x) = 3x + 2
f(x+h) = 3(x+h) + 2 = 3x + 3h + 2
f(x+h) – f(x) = 3x + 3h + 2 – (3x + 2) = 3h
[ f(x+h) – f(x) ] / h = 3h / h = 3
2.7 #60Find and for each pair of functions.
+ 3
Ans: + 3
You forgot the +3 at the end. Ans: -5x2 + 20x + 18
Ans:
The negative on the first term is not squared.
-(5x+3)2+ 4(5x+3) + 3
-(5x+3)(5x+3) + 20x + 12 + 3
- (25x2 + 15x + 15x + 9) + 20x + 15
-25x2 -10x + 6