INTRODUCTION TO RADICAL...
Transcript of INTRODUCTION TO RADICAL...
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RADICAL FUNCTIONS
INTRO TO RADICAL EXPRESSIONS Review Square Roots Extend to Cubed Roots Rationalizing the Denominator of an Expression Working with Rational Exponents ADD/SUBTRACT/MULTIPLY/DIVIDE Simplifying Radical Expressions within Operations
SOLVING RADICAL EQUATIONS Using Inverse Operations Extraneous Solutions (Check Answers) GRAPHING RADICAL FUNCTIONS Domain Transformations
APPLICATIONS
INTRODUCTION TO RADICAL EXPRESSIONS *Review β Simplify each of the following Radical Expressions. Assume all variables are positive.
1. β50 2. β32π₯5 3. β27π3π43
* Fill a value into each radical so that the result is a whole number solution, then write the solution. A couple of problems are already filled in as examples.
4. β1212
= 11 5. β____ = ________ 6. β_____2 = ________
7. β643
= 4 because 4 β 4 β 4 = 64 8. β_____3 = ________ 9 β_____4 = ________
* Review β Rationalize the denominator in each of the following. Again, the first is provided as an example.
10. π
βππ
π
βππ β
βππ
βππ =
πβππ
βππ =
πβππ
π
11. π
βππ 12.
π
βπ
* Rationalize the denominator in each.
13. π
βππ
π
βππ β
βππ
βππ =
πβππ
βππ =
πβππ
π= πβπ
π
14. Written Response: Why did the example at the left multiply by one
in the form of βππ
βππ ?
15. π
βππ 16.
ππ
βππ
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OPERATIONS WITH RADICAL EXPRESSIONS * Intro/Review β Working with Rational Exponents A Rational Exponent is a Root and a Power Combined.
Example: (8)23 means (β8
3)
2. Do you see where the 2 and the 3 from the rational exponent are in radical expression?
So (8)23 = (β8
3)
2= (2)2 = 4
*Review β Simplify each of the following Radical Expressions.
1. (27)23 2. (16)
32 3. (125)
13
*Review β Write each in radical form and then simplify. Assume all variables are positive.
4. (27π3π4)1
2 5. (ππππππ)π
π 6. (ππππππ)π
π
*Simplify each based on the mathematical operation. Assume all variables are positive.
7. (π + πβπ) β (π + βππ) 8. πβπ(πβπ + πβπ)
9. (π β πβπ)(π + πβπ)
10. βππππππβ (πππππ)
π
π β βπππ
11. βππππππ
(πππππ)ππ
12. (π + βπ)(βπ + βπ)
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*Review β *Simplify each based on the mathematical operation. Assume all variables are positive.
1. βπππ β βπππ 2. (πβπ β π) β (π β βπ) 3. βπ
βπ
4. βππππππππ β (ππππππ)π
π 5. (πβπ β π)(π β βπ) 6.
βπππ
βππππ
SOLVING RADICAL EQUATIONS Notes:
To solve a radical equation we must get rid of the radicals wrapped around the variable.
To undo a radical, square both sides of the equation after the radical is isolated.
The Steps: 1. Isolate a radical term, 2. Square both sides, 3. Continue until all radicals are gone, 4. Solve equation, 5. Check answers in original equation.
*Solve each radical equation.
1. βπ₯ β 3 = 2 2. βπ₯ + 2 = 4βπ₯ + 1 3. βπ₯2 β 16 β 3 = 0
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MIXED EXAMPLES WITH RADICAL EXPRESSIONS/EQUATIONS *Simplify each Radical Expression. Also, rationalize each denominator in necessary. Solve each equation for x.
1. β5
8 2.
3
5ββ7 3. βπ₯ β 3 β βπ₯ = 3
SOLVING RADICAL EQUATIONS * Solve each equation for the indicated unknown. Check each by graphing.
1. βπ₯ β 3 + 6 = 5 2. β9π₯2 + 4 = 3π₯ + 2 3. β4 β 2π‘ β π‘2 = π‘ + 2
4. βπ₯ + 13
β 3 = 4
MIXED EXAMPLES WITH RADICAL EXPRESSIONS/EQUATIONS *Simplify each Radical Expression. Also, rationalize each denominator in necessary. Solve each equation for x.
1. 3+5β2
4β3β10 2. β2π₯ + 5 = π₯ β 5
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* Functions Name ______________________________ Period _______
Evaluating Functions Operations with Functions Composite Functions Difference Quotient
Transformations of Radical Functions Inverse Functions Composites of Inverses
Given: π(π) = ππ + π π(π) = π + π π(π) = βπ β π π(π) = ππ + ππ π(π) = ππ β π
Evaluate each.
1. π(3)
2. π(β2) 3. β(22)
4. π(β10)
5. π(5) 6. π(β5)
Perform each operation and simplify the expression. State any domain restrictions when necessary.
7. (β + π)(π₯)
8. (π β π)(π₯) 9. (π β π)(π₯)
10. (π β β)(π₯)
11. (β
π) (π₯) 12. (
π
π) (π₯)
Evaluate and Simplify each Composite Function.
13. π(π(3))
14. β(π(5)) 15. π(π(β1))
16. (β π π)(β2)
17. (π π π)(3) 18. π(π(π₯))
19. π(π(π₯))
20. π(π(π₯)) 21. β(π(π₯))
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Extra Practice
Given: π(π) = ππ π(π) = π + π π(π) = βππ + π π(π) = ππ β ππ π(π) = ππ β π
Evaluate each.
1. π(3)
2. π(β2) 3. β(22)
4. π(β10)
5. π(5) 6. π(β5)
Perform each operation and simplify the expression. State any domain restrictions when necessary.
7. (β + π)(π₯)
8. (π β π)(π₯) 9. (π β π)(π₯)
10. (π β β)(π₯)
11. (β
π) (π₯) 12. (
π
π) (π₯)
Evaluate and Simplify each Composite Function.
13. π(π(3))
14. β(π(5)) 15. π(π(β1))
16. (β π π)(β2)
17. (π π π)(3) 18. π(π(π₯))
19. π(π(π₯))
20. π(π(π₯)) 21. β(π(π₯))
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Evaluating Piecewise Functions Given:
π(π) = {π + π ππ βπ β€ π < π
ππ β π ππ π β€ π < ππ π(π) = {
π ππ βπ β€ π < 1ππ ππ π β€ π β€ π
βπ
ππ + π ππ π > 3
π(π) = {
π β π ππ βπ < π < βππ ππ βπ β€ π < π
ππ + πβπ
ππππ
π β€ π β€ ππ > π
Evaluate each function.
1. π(π)
2. π(βπ) 3. π(π) 4. π(βπ)
5. π(π)
6. π(βπ) 7. π(π) 8. π(π)
9. π(π)
10. π(π(π)) 11. π(π(π)) 12. π (π(π(π)))
Inverses
Notes β Graphing Inverses The inverse of a function is a function that undoes the action of another function.
Graphically, and inverse is the reflection each point of the function across the line π¦ = π₯.
The line π¦ = π₯ is the set of points where every x and y value are the same, i.e., the diagonal line from lower left to upper
right.
The reflection of a point is the point that is the same distance behind the reflected surface.
Ex. 1: Reflect each point across the line π¦ = π₯.
Label the reflection of point A as Aβ, etc.
A(4, 3)
B(7, 4)
C(-5, -2)
D(-3, 6)
E(4, 4)
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Ex. 2: Graph the inverse of π¦ = 3π₯ + 1 by filling in the table for the equation, plotting the pairs of points, drawing a line
connecting these points, and then reflecting each point. Conclude by connecting the reflected points with a line.
x y
-2
-1
0
1
2
Practice
1. Graph the inverse of π¦ = 2π₯ β 4 by filling in the table for the equation, plotting the pairs of points, drawing a line
connecting these points, and then reflecting each point. Conclude by connecting the reflected points with a line.
x y
-2
-1
0
1
2
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Given: π¦ =1
2π₯ β 3 ,
a. Fill-in the table below for the equation, b. plot the pairs of points, c. draw a line connecting these points, c. reflect
each point across the line π¦ = π₯, d. Write the equation for the line created by the reflected points.
x y
-4
-2
0
2
4
Fill in the table below by using the points
from the reflected line.
x y
What do you notice when comparing this
table with the table above?
Algebraic Method of Finding the Inverse Equation
Composite of Inverse Functions
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Introduction to the Difference Quotient Donβt be afraid, but this is the basic concept of the Derivative which is used in Calculus to look at Rates of Change.
Warm-Up Evaluate and Simplify each when π(π₯) = π₯2 β 5π₯.
1. π(3)
2. π(7) 3. π(β4) 4. π(1)
5. π(π₯)
6. π(π) 7. π(π(3)) 8. π(π(β2))
9. π(π + β)
10. π(πβ1(π₯)) 11. π(π + 5) 12. π(π₯2)
You should recognize this:
π =ππ β ππ
ππ β ππ
Similar to this what is titled the Difference Quotient
π =π(π + π) β π(π)
(π + π) β π=
π(π + π) β π(π)
π
13. Evaluate and Simplify the Difference Quotient
for π(π₯) = π₯2 β 5π₯.
14. Evaluate and Simplify the Difference Quotient
for π(π₯) = 2π₯2 + 3.
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More on Domain Restrictions Notes:
Definition: Domain - the set of all possible input values (usually x), which allows the function formula to work.
(the x-values that are allowed to be used in a function)
Sometimes, rather than figuring out the x-values that are allowed, it is easier to figure out the x-values that
are not allowed and then state that you can use all x-values except for those.
Domain restrictions can occur within even roots, like square roots, and in the denominator of fractions. - We cannot take the square roots of negative numbers, therefore expressions within the square root must
be greater than or equal to 0.
- We cannot divide by zero, therefore expressions in the denominator cannot be equal to zero.
*State the domain for each of the following functions.
Ex. 1: π(π₯) = βπ₯ + 1 Ex. 2: π(π₯) =π₯+4
π₯β1 Ex. 3: π(π₯) = π₯2 β 3π₯ + 4
Practice
* State the domain for each of the following functions.
1. π(π₯) = β5π₯ β 13 2. π(π₯) =π₯+1
(3π₯β4)(π₯+2) 3. π(π₯) = 6π₯3 + 6π₯ β 1
4. π(π₯) = β5π₯ + 1 5. π(π₯) =π₯β9
(π₯β1)(2π₯+7) 6. π(π₯) = π₯2 β 3π₯ + 4
Inverses of Quadratic Functions
Algebraically determine the inverse of each quadratic function. Check for correctness using composites.
1. π(π₯) = π₯2
2. π(π₯) = (π₯ β 7)2 3. β(π₯) = π₯2 + 5
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Notes:
Keep in mind that a radical function is the inverse of a quadratic function.
You already should somewhat have a sense of this because you have learned that squaring and square rooting are inverse
operationsβ¦they undo one another.
Therefore when you see the graph of a radical function, it will look like half of a parabola on its side.
*State the domain of each radical function and then graph.
Ex. 1: π(π₯) = βπ₯ Domain: ____________ Range: ____________
x
y
This is the graph of the parent function for radical functions. You will use your understanding of transformations to accurately graph the rest.
1. Graph π¦ = βπ₯ β 3 + 4 Domain: ____________ Range: ____________
2. Graph π¦ = ββπ₯ + 2 Domain: ____________ Range: ____________
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3. Graph π¦ = 3βπ₯ Domain: ____________ Range: ____________
4. Graph π¦ = 2βπ₯ + 5 β 6 Domain: ____________ Range: ____________
5. Graph π¦ = ββπ₯ + 3 Domain: ____________ Range: ____________