College Algebra MATH 107 Final Examination Solutions - Summer 2019

This document contains the complete solutions to a College Algebra final examination (MATH 107) from the Summer of 2019. The solutions cover a wide range of topics, including function evaluation, domain and range determination, solving equations (linear, quadratic, and rational), finding the inverse of a function, and optimization problems. Each problem is solved step-by-step, demonstrating the application of key algebraic concepts. The solutions include detailed explanations and calculations, making it a valuable resource for students studying college algebra and preparing for similar exams. The exam questions cover multiple-choice, short answer, and problems requiring detailed work to be shown. Topics covered include piecewise functions, solving equations, increasing functions, diameter and center of a circle, symmetry, equations of lines, linear equations, composite functions, quadratic equations, inverse functions, and profit maximization.

Q 22.
f(x) = x − 15 and g(x) =
√ x − 3
a)
f
g (x) = x − 15
√ x − 3
[Put x=12]
∴ f
g (12) = 12 − 15
√ 12 − 3
= −3
√ 9 = −1
b)
Domain:It is the interval over which the function is defined.
The functionf
g has a square root in the denominator which is defined only
non-negative numbers.Also the function is undefined when the denominator
is zero, therefore
x − 3 > 0 =⇒ x > 3
The domain of the function
f
g is
{x ∈ R | x > 3}
Q 23.
Answer
a)
The end points of the diameter are:
(x1, y1) = (−6, 5) and (x2, y2) = (4, 3)
Diameter of the circle is given by
d =p (∆y)2 + (∆x)2 = p (y2 − y1)2 + (x2 − x1)2 = p (3 − 5)2 + (4 − (−6))2 = √ 104
b)
Center C of the circle is the midpoint of diameter, which is
C = x1 + x2
2 , y1 + y2
2 = −6 + 4
2 , 5 + 3
2 = (−1, 4)
1

c)
Symmetry about y- axis is obtained by changing the sign of x- coordinate.
Therefore, the point (1, 4) is symmetric to C about y- axis.
Q 24.
Answer
Line passes through points:(x1, y1) = (1,-7) and (x2, y2) =(3,-9).
Slope of the line is given by
m = ∆y
∆x = y2 − y1
x2 − x1
= −9 − (−7)
3 − 1 = −1
Equation of line passing through point (x1, y1) =(1,-7),having slope m =
−1, is
y − y1 = m(x − x1) =⇒ y + 7 = −1(x − 1)
Re-arranging the equation in slope-intercept form:
y = −x − 6
Q 25.
Answer
Let the annual income be x $.
Tax paid is a fixed amount of 90 $ and 1.4 % of x.Therefore,
Total tax paid = 90 +
1.4
100
x
The total tax paid for the year was 1035 $.Implies,
90 +1.4
100
x = 1035 =⇒ 1.4
100
x = 945 =⇒ x = 67500
Therefore, the annual income is 67, 500 $
Q 26.
f (x) = 4x2 − 20 and g(x) = x − 3
Answer
a)
2

(f ◦ g)(x) = 4(g(x))2 − 20
= 4(x − 3)2 − 20
= 4(x2 − 6x + 9) − 20
∴ (f ◦ g)(x) = 4x2 − 24x + 16
b)
(f ◦ g)(x) = 4x2 − 24x + 16
[Put x= -2]
(f ◦ g)(−2) = 4(−2)2 − 24(−2) + 16
= 80
Q 27.
5x2 + 12 = 20x
Answer
5x2 + 12 = 20x
[Subtract 20x from both sides]
∴ 5x 2 − 20x + 12 = 0
[Divide throughout by 5]
x2 − 4x + 2.4 = 0
The quadratic equation ax2 + bx + c = 0, has roots
x = −b ±√ b2 − 4ac
2a
Comparing,
ax2 + bx + c' x 2 − 4x + 2.4
3

we have, a = 1, b = −4 and c = 2.4.Therefore, the roots are
x = 4 ±p 42 − 4(1)(2.4)
2 × 1 = 2 ± 2
r 2
5
Therefore, the exact solutions to the given equation are
x = 2 + 2
r 2
5, x = 2 − 2
r 2
5
Q 28.
f(x) = 7 − 1
5x
Answer
We have,
y = 7 −1
5x
Interchange x and y.Therefore,
x = 7 −1
5y
[Solve for y]
1
5y = 7 − x
∴ y = 35 − 5x
[Replace y with f−1]
f −1(x) = 35 − 5x
To verify that the inverse obtained is correct, check if
(f ◦ g)(x) = x
(f ◦ g)(x) = 7 −
1
5(35 − 5x)
= 7 − 7 + x
= x =⇒ verified
4

Therefore, the inverse of f (x) is
f −1(x) = 35 − 5x
Q 29.
P (x) = −0.002x2 + 4.3x − 1400
Answer
a)
Profit when x = 850,
P (850) = −0.002(850)2 + 4.3 × 850 − 1400 = 810
Therefore, the profit made with 850 doughnuts is 810$
b)
Maximum of P (x) occurs at a stationary point which is obtained by solving
d
dxP (x) = 0
Derivative of P (x) is
d
dx(−0.002x2 + 4.3x − 1400) = −0.004x + 4.3
Therefore,
−0.004x + 4.3 = 0 =⇒ x =
4.3
0.004
= 1075
For maximum to occur at a point the required condition is that the second
derivative at that point should be negative, that is
d2
dx2 P (x) x < 0
But,
d2
dx2 P (x) = −0.004
which is always negative.
Therefore,the company should make 1075 doughnuts daily to maximize
profit.
Q 30.
5

Answer
x − 3
x + 2= 20
x2 − 4
[x2 − 4 = (x + 2)(x − 2)]
∴ x − 3
x + 2= 20
(x + 2)(x − 2)
[Multiply throughout by (x+2)]
x − 3
x + 2× (x + 2) = 20
(x + 2)(x − 2)
× (x + 2)
[If, x 6= -2, cancel(x + 2)]
∴ x − 3 = 20
x − 2
[Multiply throughout by (x − 2)]
(x − 3)(x − 2) =20
x − 2× (x − 2)
[If, x 6= 2, cancel(x − 2)]
x2 − 5x + 6 = 20
[Subtract 20 from both sides]
x2 − 5x − 14 = 0
=⇒ x 2 − 7x + 2x − 14 = 0
x(x − 7) + 2(x − 7) = 0
∴ (x − 7)(x + 2) = 0
=⇒ x = 7 or − 2
But, x = −2 is already ruled out (as the function is undefined at x = −2),
therefore, x = 7 is the solution to the given equation.
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