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Ch 4 Rational, Power, and Root Functions

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Chapter 4: Rational, Power, and Root
Functions
4.1 Rational Functions and Graphs
4.2 More on Graphs of Rational Functions
4.3 Rational Equations, Inequalities,
Applications, and Models
4.4 Functions Defined by Powers and Roots
4.5 Equations, Inequalities, and Applications
Involving Root Functions
Copyright В© 2007 Pearson Education, Inc.
Slide 4-2
4.3 Rational Equations, Inequalities,
Applications, and Models
• Solving Rational Equations and Inequalities
– at least one variable in the denominator
– may be undefined for certain values where the
denominator is 0
– identify those values that make the equation (or
inequality) undefined
– when solving rational equations, you generally multiply
both sides by a common denominator
– when solving rational inequalities, you generally get 0
on one side, then rewrite the rational expression as a
single fraction
Copyright В© 2007 Pearson Education, Inc.
Slide 4-3
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}.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-4
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.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-5
4.3 Solving a Rational Inequality
Analytically
Example
Solve the rational inequality
xпЂ«2
2x пЂ«1
п‚Ј 1.
Analytic Solution We can’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
Copyright В© 2007 Pearson Education, Inc.
пЂ­
пЂ­1п‚Ј 0
2x пЂ«1
2x пЂ«1
п‚Ј0
Common denominator is 2x + 1.
Slide 4-6
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
Copyright В© 2007 Pearson Education, Inc.
and
2x пЂ« 1 пЂЅ 0
пЂ­ 1.
2
Slide 4-7
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 ).
Can’t include  12 ; it makes the denominator 0.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-8
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
Copyright В© 2007 Pearson Education, Inc.
Slide 4-9
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
Copyright В© 2007 Pearson Education, Inc.
Slide 4-10
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 {–5}.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-11
4.3 Solving Equations Involving Rational
Functions
Solving a Rational Equation
1. Determine all values for which the rational
function is undefined.
2. Multiply both sides of the equation by the
least common denominator of the entire
equation.
3. Solve the resulting equation.
4. Reject any values that were determined in
Step 1.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-12
4.3 Applications and Models of Rational
Functions: 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) Traffic intensity x is 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) = 2 (1xпЂ­ x ) where 0 п‚Ј x <1 is
the traffic intensity. Compute f (x) for this parking ramp.
(c) Graph y = f (x). What happens to the number of vehicles
waiting as the traffic intensity approaches 1?
2
Copyright В© 2007 Pearson Education, Inc.
Slide 4-13
4.3 Applications and Models of Rational
Functions: 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
Copyright В© 2007 Pearson Education, Inc.
. 8125
2
2 (1 пЂ­ . 8125 )
п‚» 1 . 76 vehicles
Slide 4-14
4.3 Applications and Models of Rational
Functions: 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.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-15
4.3 Applications and Models 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
Two unknowns: radius x
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
Copyright В© 2007 Pearson Education, Inc.
So
hпЂЅ
2000
пЃ°x
2
.
Slide 4-16
4.3 Applications and Models 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
Copyright В© 2007 Pearson Education, Inc.
Slide 4-17
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
y пЂЅ
n
.
k
If x ,then y is inversely proportional to x, or
y varies inversely as x.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-18
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
I пЂЅ
k
d
2
models this phenomenon. At a distance of 3 meters, a 100-watt
bulb produces an intensity of .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
7.92
solve for k.
k
I пЂЅ
2
d
.8 8 пЂЅ
k
3
2
k пЂЅ 7 .9 2
Copyright В© 2007 Pearson Education, Inc.
I пЂЅ
d
I пЂЅ
2
7.92
2
2
I пЂЅ 1.98
Slide 4-19
4.3 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.
Copyright В© 2007 Pearson Education, Inc.
Slide 4-20
4.3 Solving a Combined Variation
Problem in Photography
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
Copyright В© 2007 Pearson Education, Inc.
2
16 пЂЅ F
2
4пЂЅ F
An F-stop of 4 would be
appropriate.
Slide 4-21
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