Download solutions - CSUN.edu

Math 102 Practice Final
1
Practice Math 102 Common Final Exam
1.
For A = {x|x < 1} and B = {x|x < 5}, find A1 B .
Express your answer in interval notation.
Set A is the real numbers less than 1:
Set B contains all numbers up to 5:
<))))))))))))B))))))))))))))))>
<))))))))))))))))))))))))C))))))>
A 1 B = { x | x is both < 1 and < 5 } = {x| x < 1} = A = (–4 ,1) .
A 1 B = A because A is a subset of B (everything in A is included in B).
2.
Simplify completely. Express your answer without negative exponents.
(2x4y–4/5)5(16y–4)–1/4
First we give ou rselve s a little sp ace to thin k in, then d istribute th e expo nents, 5 and –1/4:
(2 x4 y–4/5 )5 ( 16 y –4 )–1/4
=
25 x20 y–4
( 16 y –4 )–1/4
=
25 x20 y–4
16–1/4 y –4 (–1/4)
=
25 x20 y–4
16–1/4 y
=
–1/4
5
4 –1/4
2 16
20
x
Regro up to facil itate com bining li ke factor s...
–4
y y
20
x
y–4 y
=
2 (2 )
=
25 2–1 x20 y–4 y
=
24 x20 y–3
24 x20
))))))))
y3
=
3.
5
e.g. (y–4/5 )5 = y(–4/5)5 = y–4
Factor completely.
This s implific ation req uires on ly one o r two step s,
... which were drawn ou t to reveal all details
x3/2 + 6x1/2 + 9x–1/2
x1/2 (x3/2 + 6x1/2 + 9x–1/2 )
))))
x1/2
Observation: Multiplication by
x 1/2 would make that a lot nicer!
(x3/2+1/2 + 6x1/2+1/2 + 9x–1/2+1/2 )
x
)))))))))))))))))))))))))))))))))
1/2
(x2 + 6x1 + 9 )
x
))))))))))))))))
1/2
(x + 3) 2
x
)))))))))
1/2
4.
Divide as indicated. Express your answer in the form a + bi.
6i + 9
))))))
1 – 3i
=
(6i + 9)(1 + 3i)
9C1 — 6C3 + 9C3i +6i
))))))))))))) )) = ))))))))))))))))))))
(1 – 3i)(1 + 3i)
1+9
= –.9 +3.3i
Math 102 Practice Final
5.
2
Solve. Express your answer in interval notation.
x
< 2
x+4
))))))
Resist the temptation to multiply both sides.
compare to 0, not 2 !! and SIMPLIFY
x
x+4
– 2 < 0
x
x+4
2(x+4)
– ))))))
(x+4)
))))))
))))))
x – 2(x+4)
x+4
)))))))))))
< 0
–8
–x – 8
x+4
< 0
quotient
–x – 8
x+4
))))))
6.
< 0
–4
)))))) |)))))))) |)))))
+
–
–
–
–
+
–
+
–
check endpoints: at – 8: 0 < 0 (inequality is true)
at – 4: inequality false
Statement is true
for x in (–4 ,–8) U [– 4,4 )
Solve. Express your answer using interval notation.
2| (½)x + 3| + 3 < 15
I like to see the x with coefficient 1, so I
go ahead and multiply 2| (½)x + 3|
| x + 6 | + 3 < 15
|x + 6 | < 12
Here I LOOK at it and solve:
|x – –6 | < 12
... says “distance between x & –6" < 12
[ –6 – 12, –6 + 12]
[ –18 , 6 ]
But this works, too:
|x + 6 | < 12 if and only if
–12 < x + 6
–18 <
7.
x
< 12
< 6
so x is in [ –18 , 6 ]
Let A = (2,–5) and B = (–1,3) be points in the xy-plane.
a. Find the length of segment AB. (simplify the answer)
Length of AB =
)))))))))))
/ ()x)2 + ()y)2
= ( 3 2 + 8 2 )1/2
=
731/2
b. Find the midpoint of segment AB.
Midpoint of AB = (halfway between 2 & –1, halfway between –5 & 3)
=(
' 2 , ( –5
( 2 + –1 )
= ( ½ , –1 )
+ 3)
'2 )
Math 102 Practice Final
8.
3
Find the center and radius of the circle with equation ...
This is most easily done by
completing the square.
x 2 + y 2 – 4x – 6y – 3 = 0
x 2 – 4x
+ y 2 – 6y
( x – what) 2
+ ( y – what) 2
x 2 – 4x + 4
( x – 2) 2
=
+
Y2
=
Preparing to complete the sq uare
3 + whatever I need....
+ y 2 – 6y + 9
+ ( y – 3) 2
X2
= 3
= 3+4+9
3 + 13
=
16
Fill in the gaps... and
compa re to the essen tial equation of a c ircle
So we see this is the equation of a circle with center at (2,3) and radius 4.
9.
The manager of a cheap plastic furniture factory finds that it costs $2400 to produce 200
chairs in a day, and $6200 to produce 400 chairs in a day. Assuming that the
relationship between cost (y) and the number of chairs produced (x) is linear, find an
equation that relates x and y.
Reduced to its essence, this problem asks us to find the equation of a line
through (x,y) = (200,2400) and (400,6200).
SLOPE = )y/)x = 3800/200 = 19
Here I use the point-slope form:
the equation is y – y0 = 19 (x – x0)
where (x 0,y 0) is any point on the line
y – 2400 = 19 (x – 200)
y = 19x – 1400
Here I use the slope-intercept form:
to find b, I take advantage of the fact
that when x = 200, y = 2400:
10.
THIS IS SUFFICIENT
But you may also write this.
y = 19x + b
2400 = 19(200) + b
So
b = 2400 - 19(200) = -1400.
For f(x) = 2x2 – 5x + 3, find & simplify completely:
f(x+h) – f(x)
))))))))))) =
h
2(x+h)2 – 5(x+h) + 3 — ( 2x2 – 5x + 3)
h
))))))))))))))))))))))))))))))))))))))
2( x2 + 2xh + h2 ) – 5(x+h) + 3 — ( 2x2 – 5x + 3)
h
=
)))))))))))))))))))))))))))))))))))))))))))))))
=
))))))))))))))))))))))))))))))))))))))))))))))
=
=
Distribute that — !!
2x2 + 4xh + 2h2 – 5x – 5h + 3 — 2x2 + 5x — 3
h
4xh + 2h2
– 5h
)))))))))))))))))))))
h
4x + 2h – 5
... And there you have it,
in excruciating detail....
Math 102 Practice Final
11.
Sketch the graph of f(x) = r%x%%
Use transformations to sketch the graph of y =
4
–x
r%%%%%
–2
I used intermediate y = r%%–%x%%
x
12.
Find the formula for f–1(x) given f(x) below:
1
f(x) = )))))))
3x – 2
1
y = ))))))
3x – 2
We solve for x in terms of y. this will reveal the
function that, given y, finds x.
3xy – 2y = 1
3xy = 1 + 2y
1 + 2y
x = )))))))
3y
13.
We have found x in terms of y. The connection to
the inverse is... f –1(y) = x...
f–1(y) =
)))))))
1 + 2y
3y
But it does not matter what the variable is called...
f–1(x) =
)))))))
1 + 2x
3x
It’s still the same function....
Sketch the graph of P(x) = –x3 + 4x2 – 4x .
Label all intercepts and describe the end behavior.
As x 64, P(x) 6 – 4
since as x grows larger, the leading term dominates; that is, as x64,
– x3 6!4 faster than 4x2 – 4x 6 +4 , so the sum of the two 6!4.
Similarly, as x 6 – 4 , P(x) 6 + 4 .
. . . . . . . . . . . .
. . . . . . . . .
P(0) = 0, so that’s the y=intercept, at (0,0)
. . . . . . . . .
. . . . . . . . .
x-intercepts:
. . . . . . . . . . . .
. . . . . . . . . . . .
–x3 + 4x2 – 4x = 0
. . . . . . . . .
2
–x (x – 4x + 4) = 0 .
. . . . . . . . . . . .
. . . . . . . . .
–x (x – 2)(x – 2) = 0
. . . . . . . . .
. . . . . . . . .
... we see x-intercepts at 0, 2, and 2.
. . . . . . . . . . . .
. . . . . . . . .
(make allowances for the graphing...) . . . . . . . . . . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. . . .
. .
. . . .
. . . .
. . . .
. .
. .
. . . .
. .
. . . .
. . . .
. . . .
. .
. . . .
. .
.
.
.
.
.
.
.
.
Math 102 Practice Final
14.
5
For P(x) = 2x3 + 8x2 + 3x – 10 ,
a.
List all the possible rational zeros of P(x)
Because the coefficients of this polynomial are all integers, we know that any
rational roots must be of the form p/q where p divides 10, and q divides 2.
This yields: ±1, 2, 5, 10, 1/2, and 5/2
b.
Use synthetic division to show that – 2 is a zero of P(x).
–2
2
2
c.
8
–4
4
3
–8
–5
–10
10
0
This tells us P(x) = (x +2)(2x2 + 4x – 5)
So –2 is a root of P.
Find all the zeros of P(x). Simplify your answers.
The remaining roots of P must be found in (2x2 + 4x – 5)
Using the quadratic formula to solve 2x2 + 4x – 5 = 0... we find:
x =
–4 ± q %1% 6% %+
% %4
%0
%
))))))))))
4
=
–1 ±
14
2
q %% % %
)))
Of course the third root of this cubic polynomial is –2, the root we found above.
15.
Find a fourth degree polynomial P(x) with 1+2i a zero, 1 a zero of multiplicity 2, and
a constant term of 10. Express your answer in the form P(x) = ax3 + bx3 + cx2 + dx + e,
where a,b,c,d,e, are integers.
To be integers, a,b,c,d,e must all be real numbers. Since then P has only real
coefficients, any complex roots must be paired with their conjugates as roots.
Thus we know the four roots of P must be 1+2i, 1–2i, 1 and 1.
And P must be the product of these factors:
P(x) = A (x – (1+2 i) ) (x – (1–2i) ) (x – 1) (x –1)
where A is some non-zero constant.
This is:
P(x) = A (x2 – 2x + 5) (x2 – 2x + 1) which has constant term 5A.
Since they want the constant term to be 10, A had better be 2.
P(x) = 2 (x2 – 2x + 5) (x2 – 2x + 1)
P(x) = 2x4 – 4x3 + 20x2 – 24x + 10
Math 102 Practice Final
16.
6
Find the x and y intercepts. (express as ordered pairs)
Find the vertical asymptote. (write the equation)
Find the horizontal asymptote. (write the equation)
Sketch the graph.... for:
x– 6
y = ))))))
x+ 2
Domain: y is defined for all values
except x = –2
As x approaches –2, x– 6 ÿ –8,
while x+2 ÿ 0,
so the quotient ÿ ± 4
Thus there is a vertical
asymptote at x = –2.
y=1
As x ÿ4
y ÿ1
As x ÿ4
y ÿ1
Thus the function has a
horizontal asymptote at y = 1
y-intercept: when x = 0
y = –3
x-intercept: when x = 6, y = 0
17.
18.
.
. . . .
.
. . . .
.
.
.
. . . .
.
.
.
.
.
.
.
.
.
.
.
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
x= –2
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. . .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. . . . . . .
. . . . . .
. . . . . . .
. . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . .
. . . . .
.
. . . . . . .
.
. . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
. . . . . . .
(6,0)
(0,–3)
Sketch the graph of f(x) = – ex–2 (Label the asymptote and one point on the graph.)
. . . . . . . . . . . . . . . .
Working from the known graph of
. . . . . . . . . . . . . . . . . .
y = ex
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
y = ex–2 is the same graph, shifted to
. . . . . . . . . . . . . . . .
the right two units
. . . . . . . . . . . . . . . . . .
y=0 . . . . . . . . . . . . . . . .
. . . . . . . . . . (2,–1) . . . . .
y = –ex–2 is the same graph,
reflected through the x-axis
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . .
. .
. .
. .
. .
. .
.
.
.
.
.
.
.
.
.
.
.
.
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
. .
Simplify completely:
5
4 log4(2x) – log4 x + log4(4x)
log4(2x)4 – log4 x5 + log4(4x)
(2x)4 (4x)
log4 )))))))))
x5
log4 64
.
.
.
.
.
.
Method A:
Using the properties of logarithm functions,
we can combine these terms into one log
expression, and see if the result can be
simplified. It can.
Math 102 Practice Final
18.
7
Simplify completely:
Method B:
Using the properties of logarithm functions,
we can expand & separate these
terms so the log4x terms are alone.
4 log4(2x) – log4 x5 + log4(4x)
4 log42+ 4log4 x – 5log4x + log44 + log4x
log416 + log44 + (4 – 5 + 1) log4x
2 + 1
+
0
19.
Solve for x (Give an exact answer.):
ex – 2 = 3x
Std. Operating Procedure: exponential function,
ln(e x – 2) = ln(3 x )
haul out the logs....
(x – 2) ln(e) = x ln (3)
x–2
and use those logarithmic properties!
= x ln 3
... and keep in mind that ln3 is just a number.
x – xln3 = 2
( 1 – ln 3) x = 2
x = 2/(1– ln3)
20.
21.
Well, that’s an exact answer.
Solve for x:
log 6 (x–3) + log 6(x+2) = 1
log 6 ((x–3)(x+2)) = 1
((x–3)(x+2)) = 61
x2 – x – 6 = 6
x2 – x – 12 = 0
(x – 4)(x+3) = 0
x = 4 or –3
Better to have one log expression than two,
Now we can “undo” the log 6 using exponential
base 6.
CHECK THOSE SOLUTIONS!!!!
Only x = 4 is a valid solution.
$2000 is invested in an account that pays 3% (annual percentage rate)
compounded continuously. How long will it take for the balance to reach $3000?
When amount P0 is invested at r % interest compounded N times per year, the value of the account
at time t years is
Nt
P(t) = P0 ( 1 + r/N )
rt
As N 64, this approaches
P0 e
... the value achieved with continuous compounding.
.03t
So, the value of this account after t years is $2000 e
and we want to solve for t so that this value will be $3000.
.03t
$2000 e
e
.03t
.03 t
t
= $3000
= 1.5
= ln 1.5
=
ln 1.5 years Ñ 13.52 years
.03
)))))