College Algebra MATH 107 Final Examination Solution - University

This document presents a comprehensive solution to the MATH 107 final examination in College Algebra, covering a range of topics including piecewise functions, solving equations, analyzing graphs, and complex numbers. The solution provides step-by-step explanations and answers to multiple-choice and short-answer questions. The assignment includes detailed solutions for various problems, such as determining the domain and range of functions, solving equations, identifying intervals of decreasing functions, and working with complex numbers. It also covers topics like finding the vertex of a parabola, determining the range of a function, and analyzing the critical points and intervals of increasing/decreasing behavior of polynomial functions. Additionally, the solution demonstrates composite functions, domains, and y-intercepts, along with analyzing the graph of a polynomial function.

Running head: MATH 107 FINAL EXAMINATION
MATH 107 FINAL EXAMINATION
Name of the Student
Name of the University
Author Note

1MATH 107 FINAL EXAMINATION
3.
From the graph of the function it can be seen that the function is decreasing in (-∞,-1) and
(3,∞) and in (-2.5,1) and (4.5,∞).
Hence, correct options are C and D.
4.
y = 3|x| (1)
Replacing y with –y in (1) gives
-y = 3|x|. This is not equivalent to (1) hence the graph is not symmetric about x axis.
Replacing x with –x in (1) gives
y = 3|-x| => y = 3|x|.
This is same as (1) and hence the graph is symmetric about y axis.
Replacing y =-y and x = -x in (1) gives
-y = 3|-x| => -y = 3|x|
This is not same as (1) and hence graph is not symmetric about the origin.
Hence, option (B) is correct.
8.
By vertical line test the function is a graph of a function as no vertical line intersects graph at
more than one point. By horizontal line test there exist more than one point where horizontal
line intersects graph and hence the function is not one to one. The graph is not parabolic.
Hence, option D is correct.

2MATH 107 FINAL EXAMINATION
12.
Shifting y=f(x) one unit on right and then shifting 2 units downward gives y = g(x).
Hence, option B is correct.
13.
(5+8i)(3+i) = 15 + 5i + 24i -8 = 7 + 29i
a = 7 and b = 29.
19.
f(x) = x^2 -14x + 51
a) y = x^2 – 2*7x +(-7)^2 + 2 = (x-7)^2 + 2
=> y= ( x −7 ) 2 +2
Hence, the vertex of the parabola is (7,2)
b) The range of the function is the y values.
Range = [2,∞)
c) The critical point of the function is
f’(x) = 2x -14
 2x -14 = 0
 x = 7
Now, when x > 7 suppose x = 10
f’(10) = 2*10 – 14 = 6.
Hence, in (7,∞) the function is increase.

3MATH 107 FINAL EXAMINATION
Now, when x<7, suppose x = 0
f’(0) = 2*0 – 14 = -14 < 0
Hence, the function is decreasing in the interval (-∞,7).
20.
P(x) = -(1/10)x^4 – (½)x^3 + (7/10)x^2 + (41/10)x + 3
=> P’(x) = -(2/5)x^3 – (3/2)x^2 + (7/5)x + 41/10
Now, critical points are
-(2/5)x^3 – (3/2)x^2 + (7/5)x + 41/10 = 0
=> x = 1.725, -1.492 and -3.98
Now, when x > 1.725 suppose for x = 2
P’(2) = -(2/5)*2^3 – (3/2)*2^2 + (7/5)*2 + 41/10 = -2.3.
Hence, the function is decreasing towards the right end
Now, when x < -3.98 suppose for x = -4
P’(-4) = 0.1
Hence, the function is increasing in the left end.
Thus option D is correct.
b) The zeroes of the function are x = 3, x = -1, x = -2 and x = -5.
c) The y-intercept of the function is found by putting x = 0 in the equation.
y = 3
Hence, the y-intercept is (0,3)

4MATH 107 FINAL EXAMINATION
d) Graph A is correct graph of P(x).
22.
f(x) = √ x+2 and g(x) = x-3
a) ( f
g ) ( x ) = √ x +2
x−3
Hence, ( f
g ) (−1 ) = √−1+2
−1−3 = -1/4
b) Domain of f
g is the domain of g.
Domain of g is {x ϵ R∨x ≠ 3 } as g(x) not defined at x = 3.
26.
f(x) = 4x^2 + 9 and g(x) = x+3
a) Composite function (f ∘ g)(x) = 4∗( x+3 )2+9
b) (f ∘ g)(-5) = 4∗( −5+3 ) 2 +9 = 25.