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# hpl05_0403

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```4.3 Rational Equations, Inequalities,
Applications, and Models
вЂў Solving Rational Equations and Inequalities
вЂ“ at least one variable in the denominator
вЂ“ may be undefined for certain values where the
denominator is 0
вЂ“ identify those values that make the equation (or
inequality) undefined
вЂ“ when solving rational equations, you generally multiply
both sides by a common denominator
вЂ“ when solving rational inequalities, you generally get 0
on one side, then rewrite the rational expression as a
single fraction
Slide 4.3-1
4.3 Solving a Rational Equation
Analytically
Example
Solve
xпЂ«2
2x пЂ«1
пЂЅ 1.
Analytic Solution Notice that the expression is
undefined for x пЂЅ пЂ­ 12 .
xпЂ«2
2x пЂ« 1
пЂЅ1
x пЂ« 2 пЂЅ 1( 2 x пЂ« 1)
Multiply both sides by 2x + 1.
x пЂ« 2 пЂЅ 2x пЂ« 1
1пЂЅ x
Solve for x.
The solution set is {1}.
Slide 4.3-2
4.3 Solving a Rational Equation
Analytically
Graphical Solution
xпЂ«2
2x пЂ« 1
Rewrite the equation as
пЂ­ 1 пЂЅ 0 and define Y1 пЂЅ
xпЂ«2
2x пЂ«1
пЂ­ 1.
Using the
x-intercept method shows that the zero of the function
is 1.
Slide 4.3-3
4.3 Solving a Rational Equation
x
пЂ«
1
пЂЅ
8
Example
Solve
Solution
For this equation, x п‚№ п‚± 2 .
xпЂ­2
x
xпЂ­2
пЂ«
xпЂ«2
1
xпЂ«2
пЂЅ
x пЂ­4
2
.
8
x пЂ­4
2
x ( x пЂ« 2 ) пЂ« 1( x пЂ­ 2 ) пЂЅ 8
x пЂ« 2x пЂ« x пЂ­ 2 пЂЅ 8
2
x пЂ« 3 x пЂ­ 10 пЂЅ 0
2
Slide 4.3-4
4.3 Solving a Rational Equation
( x пЂ« 5 )( x пЂ­ 2 ) пЂЅ 0
x пЂЅ пЂ­5
or
xпЂЅ2
But, x = 2 is not in the domain of the original
equation and, therefore, must be rejected. The
solution set is {вЂ“5}.
Slide 4.3-5
4.3 Solving a Rational Inequality
Analytically
Example
Solve the rational inequality
xпЂ«2
2x пЂ«1
п‚Ј 1.
Analytic Solution We canвЂ™t multiply both sides
by 2x + 1 since it may be negative. Start by
subtracting 1 from both sides.
xпЂ«2
2x пЂ«1
xпЂ«2
2x пЂ«1
пЂ­
пЂ­1п‚Ј 0
2x пЂ«1
2x пЂ«1
п‚Ј0
Common denominator is 2x + 1.
Slide 4.3-6
4.3 Solving a Rational Inequality
Analytically
x пЂ« 2 пЂ­ ( 2 x пЂ« 1)
2x пЂ«1
x пЂ« 2 пЂ­ 2x пЂ­1
2x пЂ«1
пЂ­ x пЂ«1
2x пЂ«1
п‚Ј0
Rewrite as a single fraction.
п‚Ј0
п‚Ј0
To determine the sign graph, solve the equations
пЂ­ x пЂ«1пЂЅ 0
to get x = 1 and x пЂЅ
and
2x пЂ« 1 пЂЅ 0
пЂ­ 1.
2
Slide 4.3-7
4.3 Solving a Rational Inequality
Analytically
Complete the sign graph and determine the intervals
where the quotient is negative.
The quotient is zero or negative when x is in ( пЂ­п‚Ґ , пЂ­ 12 ) пЃ• [1, п‚Ґ ).
CanвЂ™t include пЂ­ 12 ; it makes the denominator 0.
Slide 4.3-8
4.3 Solving a Rational Inequality
Graphically
Graphical Solution Let Y1 пЂЅ
xпЂ«2
2x пЂ«1
пЂ­ 1 . We use the
graph to find the intervals where Y1 is below the
x-axis, including the x-intercepts, where Y1 = 0.
The solution set is ( пЂ­п‚Ґ , пЂ­ 1 ) пЃ• [1, п‚Ґ ).
2
Slide 4.3-9
4.3 Solving Equations Involving Rational
Functions
Solving a Rational Equation
1. Rewrite the inequality, if necessary, so that 0 is on one
side and there is a single rational expression on the
other side.
2. Determine the values that will cause either the
numerator or the denominator of the rational
expression to equal 0. These values determine the
intervals on the number line to consider.
3. Use the test value from each interval to determine
which intervals form the solution set. Be sure to
check endpoints.
Slide 4.3-10
4.3 Models and Applications of Rational
Functions: Analyzing Traffic Intensity
Example Vehicles arrive randomly at a parking ramp at an
average rate of 2.6 vehicles per minute. The parking attendant
can admit 3.2 cars per minute. However, since arrivals are
random, lines form at various times.
(a) The traffic intensity x is defined as the ratio of the
average arrival rate to the average admittance rate.
Determine x for this parking ramp.
(b) The average number of vehicles waiting in line to enter
the ramp is modeled by f (x) = x 2 where 0 п‚Ј x <1 is
,
(1пЂ­ x )for this parking ramp.
the traffic intensity. Compute f2(x)
(c) Graph y = f (x). What happens to the number of vehicles
waiting as the traffic intensity approaches 1?
Slide 4.3-11
4.3 Models and Applications of Rational
Functions: Analyzing Traffic Intensity
Solution
(a) Average arrival rate = 2.6 vehicles/min,
average admittance rate = 3.2 vehicles/min, so
xпЂЅ
2 .6
пЂЅ . 8125 .
3 .2
(b) From part (a), the average number of vehicles
waiting in line is f(.8125).
f (. 8125 ) пЂЅ
. 8125
2
2 (1 пЂ­ . 8125 )
п‚» 1 . 76 vehicles
Slide 4.3-12
4.3 Models and Applications of Rational
Functions: Analyzing Traffic Intensity
(c) From the graph below, we see that as x
approaches 1, y = f (x) gets very large, that is, the
number of waiting vehicles gets very large.
Slide 4.3-13
4.3 Models and Applications of Rational
Functions: Optimization Problem
Example A manufacturer wants to construct cylindrical
aluminum cans with volume 2000 cm3 (2 liters). What radius
and height will minimize the amount of aluminum used?
What will this amount be?
Solution
and height h. To minimize the amount
of aluminum, we minimize the surface
area. Volume V is
V пЂЅпЃ°x h
2
2000 пЂЅ пЃ° x h .
2
So
hпЂЅ
2000
пЃ°x
2
.
Slide 4.3-14
4.3 Models and Applications of Rational
Functions: Optimization Problem
Surface area S = 2пЃ°xh + 2пЃ°x2, x > 0 (since x is the radius), can
now be written as a function of x.
пѓ¦ 2000 пѓ¶
2
S ( x ) пЂЅ 2пЃ° x пѓ§
пЂ«
2
пЃ°
x
пѓ·
2
пЃ°
x
пѓЁ
пѓё
4000
2
пЂЅ
пЂ« 2пЃ° x
x
3
4000 пЂ« 2пЃ° x
пЂЅ
x
Minimum radius is approximately 6.83 cm and the height
associated with that is пЃ° (2000
п‚»13.65 cm, giving a minimum
6 . 83 )
amount of aluminum of 878.76 cm3.
2
Slide 4.3-15
4.3 Inverse Variation
Inverse Variation as the nth Power
Let x and y denote two quantities and n be a positive
number. Then y is inversely proportional to the nth
power of x, or y varies inversely as the nth power of
x, if there exists a nonzero number k such that
y пЂЅ
k
x
If y пЂЅ
k
,
n
.
then y is inversely proportional to x, or y
x
varies inversely as x.
Slide 4.3-16
4.3 Modeling the Intensity of Light
The intensity of light I is inversely proportional to the second
power of the distance d. The equation
k
I пЂЅ
d
2
models this phenomenon. At a distance of 3 meters, a 100-watt
bulb produces an intensity of 0.88 watt per square meter. Find
the constant of variation k, and then determine the intensity of
the light at a distance of 2 meters.
Substitute d = 3, and I = .88 into the variation equation, and
solve for k.
k
7.92
I пЂЅ
2
d
0.88 пЂЅ
k
3
2
k пЂЅ 7.92
I пЂЅ
d
I пЂЅ
2
7.92
2
2
I пЂЅ 1.98
Slide 4.3-17
4.3 Joint Variation
Joint Variation
Let m and n be real numbers. Then z varies jointly as the nth
power of x and the mth power of y if a nonzero real number k
exists such that
z = kxnym.
Slide 4.3-18
4.3 Solving a Combined Variation Problem
In the photography formula L пЂЅ
25 F
2
st
the luminance L (in foot-candles) varies directly as the square
of the F-stop F and inversely as the product of the file ASA
number s and the shutter speed t. The constant of variation is
25.
Suppose we want to use 200 ASA file and a shutter speed of
1/250 when 500 foot candles of light are available. What would
be an appropriate F-stop?
500 пЂЅ
25 F
2
пѓ¦ 1 пѓ¶
200 пѓ§
пѓ·
2
5
0
пѓЁ
пѓё
400 пЂЅ 25 F
2
16 пЂЅ F
2
4пЂЅ F
An F-stop of 4 would be
appropriate.
Slide 4.3-19
4.3 Rate of Work
Rate of Work
If 1 task can be completed in x units of time, then the
rate of work is 1/x task per time unit.
Slide 4.3-20
4.3 Analyzing Work Rate
Example It takes machine B one hour less to complete a task
when working alone than it takes machine A
working alone. If they start together, they can
complete the task in 72 minutes. How long does it
take each machine to complete the task when
working alone?
Solution Let x represent the number of hours it takes
machine A to complete the task alone. Then it
x пЂ­ 1 пЂ© hours working alone.
2 takes machine B пЂЁ
V пЂЅпЃ°x h
2000
2
2000 пЂЅ пЃ° x h . So h пЂЅ
.
2
пЃ°x
Slide 4.3-21
4.3 Analyzing Work Rate
Solution
1
пЂ«
1
пЂЅ
5
x пЂ­1 6
1 пѓ№
пѓ¦5пѓ¶
6 x пЂЁ x пЂ­ 1пЂ©
пЂ«
пЂЅ 6 x пЂЁ x пЂ­ 1пЂ© пѓ§ пѓ·
пѓЄ x x пЂ­1пѓє
пѓ«
пѓ»
пѓЁ6пѓё
6 пЂЁ x пЂ­ 1пЂ© пЂ« 6 x пЂЅ 5 x пЂЁ x пЂ­ 1пЂ©
x
6x пЂ­ 6 пЂ« 6x пЂЅ 5x пЂ­ 5x
2
0 пЂЅ 5 x пЂ­ 17 x пЂ« 6
2
5 x пЂ­ 2 пЂЅ 0 or
2
xпЂЅ
or
5
xпЂ­3пЂЅ 0
xпЂЅ3
The only value that makes sense is 3. It takes machine A 3
hours to complete the task alone, and it takes machine B 2
hours to complete the task alone.