06-a final exam review-2academics.cehd.umn.edu/robertson/1-1051/l43-mon-12... · 12/12/2016 ·...
TRANSCRIPT
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 27
L43-Mon-12-Dec-2016-Rev-Cpt-4-HW44-and-Rev-Cpt-5-for-Final-HW45
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 28
1. y-intercept:
2
2
3 4
0 0 3 0 40
f x x x x
f x
x-intercepts:
2
2
3 4
0 3 4 so -intercepts at 0, 3, and 4
f x x x x
x x x x
2. Multiplicities:
0, multiplicity 2, which is even so graph TOUCHES -axis here3, multiplicity 1, which is odd so graph CROSSES -axis here4, multiplicity 1, which is odd so graph CROSSES -axis here
x xx xx x
3. End behavior: Graph looks like nf x ax .
2 4,As x f x x x x x , which looks like 4y x .
4. Max number of turning points = n – 1 = 4 - 1 = 3.
5. Behavior near each intercept:
Near x = -4,
2
2 2
0 variable
3 4
3 4 3 4 112 44 4 4 84 4 44
f x x x x
f j x xx
Near x = 0,
0 variab
2 2 2
le
2 3 4
0 3 40 0 0 0 3 0 4 12
f x x x x
f g xx x
Near x = 3,
2
2 2
0 variable
3
3 4
3 4 3 4 633 3 3 63 3 3 3 189
f x x x x
f h x x xx
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 29
Plot some specific points if necessary:
2
2
2 2 2 3 2 4 40
2 2 2 3 2 4 24
f
f
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L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 31
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 32
1. Factor, reduce, find holes:
23 2
2
121255
4 35
x x xx x xx xx x
x x xx x
Domain: | 0, 5x x x
Reduce:
x
R x 4 3x xx
54 3
5
xx x
Q xx
Since the x's cancel, there will be a hole at x = 0.
To find the y-value of the hole, find
0 4 0 3 120 2.450 5
Q
So, there is a hole at (0, -2.4).
2. Intercepts
x-intercepts: These are the values of x in numerator that make function 0.
4 35
4 30
50 4 3
4 3
x xQ x
xx x
xx x
x or x
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 33
x = 4 has multiplicity 1, which is odd, so graph CROSSES here
Calculate behavior near x = 4:
0 Variable
4 35
4 3 4 34 4 74
45 5
44
4 9
x xQ x
x
Q g x xx
x = -3 has multiplicity 1, which is odd, so graph CROSSES here
Calculate behavior near x = -3:
0 Variable
4 35
4 33 3 33
43
233
537
5
x xQ x
x
Q h xx
x
y-intercept
This is R(0) but 0 is not in domain so there is no y-intercept. There is a hole at (0, -2.4)
3. Vertical asymptotes: These are values of x in denominator that make it zero.
The multiplicity of x = -5 is odd, so near –5 the graph will go up on one side and down on other.
4. Horizontal or oblique asymptotes:
Since the degree of the numerator is exactly 1 more than the degree of the denominator, there is an oblique asymptote. Use long division to find it.
2
2
6
5 12
5
6 126 30
18
x
x x x
x x
xx
OA is y = x - 6
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 34
5. Plot where graph crosses HA or OA by solving
2
2
2 2
Function oblique asymptote
12 65
12 6 5
12 3012 30
Q x y
x x xx
x x x x
x x x x
No solution so graph does not cross asymptote.
6. Plot a point or two to help with the sketch.
8 4 8 38
8 520
8 4 8 38
8 53.4
Q
Q
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1. Factor, reduce to find holes: There is a hole at x = -2. This implies a factor of x + 2 in both numerator and denominator.
22
xx
2. Intercepts:
x-int at x = -7 crosses suggests (x + 7) in numerator.
x-int at x = 5 crosses suggests (x - 5) in numerator.
722
5x xxx
y-intercept is at about y = 1.5 We can check this later.
3. Vertical Asymptote: Caused by zeros in denominator.
VA of x = -4 suggests (x + 4) in denominator.
VA of x = 3 suggests (x - 3) in denominator.
4 3
2 7 52
x x xx x x
4. Horizontal or Oblique Asymptote: Caused by relative degrees of numerator and denominator.
HA of y = ½ suggests the degree of numerator and denominator are the same and the ratio of the coefficients of the dominant terms is ½.
12
2 7 52 4 3
x x xx x x
5. Does function cross HA or OA? We cannot tell from the graph.
Check y-intercept:
0 0 0
00 0 0
1 2 7 5 7 0 1 .54 82 2 4 3
f
Checks.
Check degrees of numerator and denominator. They are the same (degree 3). Checks
Check if function crosses HA.
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 36
1 2
f x HA
x
7 5
2 2
x x
x
2
2
2 2
2 2
124 3
2 35 122 12
12 35 2 122
2 35 1223
x x
x xx x
x x x x
x x x xx
So, the graph crosses the HA when x = 23. We cannot tell from the given graph if that is so but it certainly looks possible. Graphing our function and then zooming in around x = 23 shows that the function does cross the HA there (hard to see!).
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 37
Put everything on one side and zero on the other and figure out the signs:
1 1 02 2 5
1 5 1 2 2 02 2 5 5 2 2
5 2 2 02 2 5 2 5 2
5 2 2 02 2 5
3 02 2 5
x xx x
x x x xx x
x x x xx x
x xx
x x
The zeros of all the factors are: 1 and 5 and 3x x x
Interval: -¥ -5 -1 3 ¥
Try x -6 -2 0 4
Signs of
32 2 5
xx x
PNN
PNP
PPP
NPP
Sign of f x P N P N
0f x ? yes no yes no
So, the solution is , 5 1,3 Note that x = -5 and x = -1 cause division by 0 so they are not
included in solution. x = 3 is OK since 0 in the numerator is allowed because this is not a strict inequality.
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1 1 12 212 1
xf g x f g x fxx x
xx
Domain f: Domain g: Domain 2
xx
2 1 02 1
12
xx
x
0x 2 02
xx
So 12
x 0x 2x
Domain of f g x must respect domains of g and 2
xx
. So, we get | 0, 2x x x
Note also that since 12
x for f(x) we can say 1 1 1so so 22 2
g x xx
Exchange x and y and solve for y.
1
3 4:2
3 4:2
2 3 42 3 43 2 43 2 4
2 43
xf x yxyf x x
yx y yxy x yxy y xy x x
xyx
So, the inverse function is 1 2 43
xf xx
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 39
Domain of is | 2 or ,2 2,f x x x
1Range of Domain of , which is 3f x f x x so range is
| 3 or , 3 3,y y
L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 40
We can write the exponential functions with the same base and then equate the exponents:
2
2
2
2
82
823
86
8 6
2
2
15125
155
155
5 5
8 6
6 8 02 4 0
2 or 4
xx
xx
xx
x x
x xx x
x xx x
These are both in the domain of the original equation.
Or, if we could not write the expressions with the same bases, we would use logs:
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Put all logs together on one side of equation and then change to exponential:
5 5
5 5
51
2
2
log 3 1 log 1log 3 log 1 1
log 3 1 1
3 1 5
2 3 5
2 8 04 2 0
4 or 2
x xx x
x x
x x
x xx x
x xx x
-4 is not in the domain of the original equation so the only solution is x = 2.
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It is tempting to take the log of both sides in order to get the x out of the exponent.
ln 81 2 ln 3 9x x
We can break up the right side using the product property but we cannot break up the left side.
ln 81 2 ln 3 ln 9
ln 81 2 ln 3 ln 9
x x
x x
Can we make the bases of the exponentials the same? Notice that 81 can be written as 29 .
(We could also use a base 3 since 2 49 3 and 81 3 . That would work but it would involve solving
a degree 4 equation, which is more difficult, but doable in this case since it would be reducible to a quadratic equation.)
2
2
2
81 2 3 9
9 2 3 9
9 2 3 9
9 2 3 9
x x
x x
x x
x x
Now, we have 9x and 29x so try a u substitution:
2
2
2
9 2 3 9
Let 9
2 3
3 2 02 1 0
2 or 1
x x
xuu u
u uu u
u u
Now, back substitute:
0
1
9 1
9 90
x
x
u
x
2
9 2
ln9 ln2ln9 ln2
ln2 0.6931 0.3154ln9 2.1972
x
x
u
x
x
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