N th Roots and Rational Exponents What you should learn: Evaluate nth roots of real numbers using both radical notation and rational exponent notation.

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Presentation transcript:

n th Roots and Rational Exponents What you should learn: Evaluate nth roots of real numbers using both radical notation and rational exponent notation Evaluate the expression. 6.1 nth Roots and Rational Exponents Solving Equations.

Using Rational Exponent Notation Rewrite the expression using RATIONAL EXPONENT notation. 6.1 nth Roots and Rational Exponents Ex)  If n is odd, then a has one real nth root:  If n is even and a = 0, then a has one nth root:  If n is even and a > 0, then a has two real nth roots:  If n is even and a < 0, then a has NO Real roots: simplify

6.1 nth Roots and Rational Exponents Using Rational Exponent Notation Rewrite the expression using RADICAL notation. Ex)

6.1 nth Roots and Rational Exponents Evaluating Expressions Evaluate the expression. Ex)

6.1 nth Roots and Rational Exponents Solving Equations Ex) When the exponent is EVEN you must use the Plus/Minus When the exponent is EVEN you must use the Plus/Minus When the exponent is ODD you don’t use the Plus/Minus When the exponent is ODD you don’t use the Plus/Minus

6.1 nth Roots and Rational Exponents Solving Equations Ex) 4 4 Very Important 2 answers ! Take the Root 1 st.

Evaluate the expressions. 6.1 nth Roots and Rational Exponents

Properties of Rational Exponents 6.2 Properties of Rational Exponents

Review of Properties of Exponents from section 5.1 a m * a n = a m+n (a m ) n = a mn (ab) m = a m b m a -m = = a m-n = Page 420 in your book

Ex: Simplify. (no decimal answers) a. 6 1/2 * 6 1/3 = 6 1/2 + 1/3 = 6 3/6 + 2/6 = 6 5/6 b. (27 1/3 * 6 1/4 ) 2 = (27 1/3 ) 2 * (6 1/4 ) 2 = (3) 2 * 6 2/4 = 9 * 6 1/2 ** All of these examples were in rational exponent form to begin with, so the answers should be in the same form! c. d.

Ex: Simplify. a. = = = 5 b. = = = 2 Ex: Write the expression in simplest form. a. = = = b. = = = = Can’t have a Radical in the basement! ** If the problem is in radical form to begin with, the answer should be in radical form as well

Ex: Perform the indicated operation a. 5(4 3/4 ) – 3(4 3/4 ) = 2(4 3/4 ) b. = = = c. = If the original problem is in radical form, the answer should be in radical form as well. If the problem is in rational exponent form, the answer should be in rational exponent form.

6.2 Rational Exponents and Radical Functions Simplify the Expressions Ex)

More Examples a. b. c. d.

Directions: Simplify the Expression. Assume all variables are positive. a. b.(16g 4 h 2 ) 1/2 = 16 1/2 g 4/2 h 2/2 = 4g 2 h c. d.

6.2 Properties of Rational Exponents Yes, Change both to rational exponent form and use the quotient property.

6.3 Power Functions and Functions Operations A2.2.5

Operations on Functions: for any two functions f(x) & g(x) 1.A ddition h(x) = f(x) + g(x) 2.S ubtraction h(x) = f(x) – g(x) 3.M ultiplication h(x) = f(x)g(x) 4.D ivision h(x) = f(x) / g(x) 5.C omposition h(x) = f(g(x)) OR g(f(x)) ** D omain – all real x-values that “make sense” (i.e. can’t have a zero in the denominator, can’t take the even nth root of a negative number, etc.)

Example: Let f(x) = 3x 1/3 & g(x) = 2x 1/3 F ind (a) the sum, (b) the difference, and (c) the domain for each. (a) 3 x 1/3 + 2 x 1/3 = 5 x 1/3 (b) 3 x 1/3 – 2 x 1/3 = x 1/3 (c) Domain of (a) all real numbers Domain of (b) all real numbers The SUM The DIFFERENCE

Ex : Let f(x) = 4x 1/3 & g(x) = x 1/2. Find (a) the product, (b) the quotient, and (c) the domain for each. a.) The PRODUCT b.) The QUOTIENT

Ex : Let f(x) = 4x 1/3 & g(x) = x 1/2. Find (a) the product, (b) the quotient, and (c) the domain for each. (a)4x 1/3 * x 1/2 = 4x 1/3+1/2 = 4x 5/6 (b) = 4x 1/3-1/2 = 4x -1/6 = Domain of (a) all reals ≥ 0, because you can’t take the 6 th root of a negative number. Domain of (b) all reals > 0, because you can’t take the 6 th root of a negative number and you can’t have a denominator of zero.

Composition f(g(x)) means you take the function g and plug it in for the x-values in the function f, then simplify. g(f(x)) means you take the function f and plug it in for the x-values in the function g, then simplify. 6.3 Perform Function Operations and Composition Cont’

You purchase a baseball glove with a price tag of $180 dollars. The sports store applies a newspaper coupon of $50 and a 10% store discount. When the coupon is applied before the discount. When the discount is applied before the coupon. A B We will visit this question later …

Ex: Let f(x) = 2x -1 & g(x) = x Find (a) f(g(x)), (b) g(f(x)), (c) f(f(x)), and (d) the domain of each. (a) 2(x 2 -1) -1 = (b) (2x -1 ) 2 -1 = 2 2 x = (c) 2(2x -1 ) -1 = 2(2 -1 x) = (d) D DD Domain of (a) all reals except x = ±1. Domain of (b) all reals except x = 0. Domain of (c) all reals except x = 0, because 2x-1 can’t have x = 0. g(f(x)) f(f(x))f(g(x))

You purchase a baseball glove with a price tag of $180 dollars. The sports store applies a newspaper coupon of $50 and a 10% store discount. Find the final price of the purchase when the coupon is applied before the discount. Find the final price when the discount is applied before the coupon. A B

STEP 1 Write functions for the discounts. Function for $50 coupon: f(x) = x - 50 Function for 10% discount: g(x) = x – 0.10x = 0.90x STEP 2 Compose the functions $50 coupon is applied first: g(f(x)) = g(x – 50) = 0.90(x - 50) 10% discount is applied first: f(g(x)) = f(0.90x) = 0.90x - 50

STEP 3 Evaluate the functions g(f(x)) f(g(x)) when x = 180 $50 coupon is applied first: g(f(180)) = 0.90(180 – 50) = $117 10% discount is applied first: f(g(180)) = 0.90(180) – 50 = $112 The final price is $117 when the $50 coupon is applied before the 10% discount. The final price is $112 when the 10% discount is applied before the $50 coupon.

#21) #29)

How is the composition of functions different form the product of functions? The composition of functions is a function of a function. The output of one function becomes the input of the other function. The product of functions is the product of the output of each function when you multiply the two functions. 6.3 Power Functions and Functions Operations

Inverse Functions What you should learn: Find inverses of linear functions. Verify that f and g are inverse functions. 6.4 Inverse Functions Michigan Standard A2.2.6 Graph the function f. Then use the graph to determine whether the inverse of f is a function.

Temperature Conversion The formula to convert temperatures from degrees Celsius-to-Fahrenheit is… But, how do you convert from Fahrenheit-to-Celsius?? C = Toon in later…

Review from chapter 2 Relation – a mapping of input values (x-values) onto output values (y-values). Here are 3 ways to show the same relation. y = x 2 x y Equation Table of values Graph

Inverse relation – just think: switch the x & y -values. x = y 2 x y * the inverse of an equation : switch the x & y and solve for y. ** the inverse of a table : switch the x & y. ** the inverse of a graph : the reflection of the original graph in the line y = x.

To find the inverse of a function: 1.Change the f(x) to a y. 2.Switch the x & y values. 3.Solve the new equation for y. ** Remember functions have to pass the vertical line test!

Ex: Find an inverse of f(x) = -3x+6. - switch x & y -solve for y y = -3x + 6 x = -3y + 6 x - 6 = -3y - change f(x) with y 3 Steps:

Verify Inverse Functions Given 2 functions, f(x) & g(x), if f(g(x)) = x AND g(f(x)) = x, then f(x) & g(x) are inverses of each other. Symbols: f -1 (x) means “f inverse of x”

Ex: Verify that f(x)= -4x+8 and g(x) = -1 / 4 x + 2 are inverses. Meaning find f(g(x)) and g(f(x)). If they both equal x, then they are inverses. f(g(x))= -4(- 1 / 4 x + 2) + 8 = x – = x g(f(x))= - 1 / 4 (-4x + 8) + 2 = x – = x ** Because f(g(x))= x and g(f(x)) = x, they are inverses.

Temperature Conversion The formula to convert temperatures from degrees Celsius-to-Fahrenheit is… But, how do you convert from Fahrenheit-to-Celsius?? C =

Temperature Conversion Solve for C - 32

Ex: (a) Find the inverse of f(x) = x 5. 1.y = x 5 2.x = y 5 3. (b) Is f -1 (x) a function? (hint: look at the graph! Does it pass the vertical line test?) Yes, f -1 (x) is a function.

Horizontal Line Test Used to determine whether a function’s inverse will be a function by seeing if the original function passes the h hh horizontal line test. If the original function p pp passes the horizontal line test, then its i ii inverse is a function. If the original function d dd does not pass the horizontal line test, then its i ii inverse is not a function.

Ex: Graph the function f(x)=x 2 and determine whether its inverse is a function. Graph does not pass the horizontal line test, therefore the inverse is not a function.

Ex: f(x)=2x Determine whether f -1 (x) is a function, then find the inverse equation. f -1 (x) is not a function. y = 2x x = 2y x + 4 = 2y 2 OR, if you fix the tent in the basement…

Ex: g(x)=2x 3 Inverse is a function! y = 2x 3 x = 2y 3 OR, if you fix the tent in the basement…

Finding Domain and Range State the domain and range of the function. 1) 2) All real #’s

Solving Radical Equations What you should learn: Solve equations that contain Radicals. 6.6 Solving Radical Equations Solve equations that contain Rational exponents. Michigan Standard L1.2.1

Solve the equation. Check for extraneous solutions. check your solutions!! Solve equations that contain Radicals Ex.1) Key Step: To raise each side of the equation to the same power. 6.6 Solving Radical Equations Simple Radical

6.6 Solving Radical Equations Ex.2) Key Step: isolate Before raising each side to the same power, you should isolate the radical expression on one side of the equation. Simple Radical

6.6 Solving Radical Equations One Radical Ex.3)

6.6 Solving Radical Equations Two Radicals Ex.4)

6.6 Solving Radical Equations Radicals with an Extraneous Solution What is an Extraneous Solution? … is a solution to an equation raised to a power that is not a solution to the original equation. Example 5)

6.6 Solving Radical Equations Radicals with an Extraneous Solution Ex.5)

6.6 Solving Radical Equations Radicals with an Extraneous Solution Ex.5)

Reflection on the Section Without solving, explain why 6.6 Solving Radical Equations has no solution.

6.6 Solving Radical Equations Solve equations that contain Rational exponents. Ex. 6) it

6.6 Solving Radical Equations Solve equations that contain Rational exponents. Ex. 7) it

Reflection on the Section Without solving, explain why 6.6 Solving Radical Equations has no solution.