06-a Final Exam Review-2academics.cehd.umn.edu/robertson/1-1051/L43-Mon-12... · 12/12/2016  · L43-Mon-12-Dec-2016-Rev-Cpt-4-for-Final-HW44-and-Rev-Cpt-5-for-Final-HW45 Page 28

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

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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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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

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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

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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.

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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!).

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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

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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

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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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