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toc =Chapter 8: Inverses and Radicals= Chapter 8 Preview: [|aa Chapter 8 Preview.pdf]

Wiki Summary Assignments
8-1: Abi B. and Tricia L. 8-2: Andrew E. and Josh H. 8-3: Brittani H. and Amanda T. 8-4: Isaiah C. and Jake L. 8-5: Paige H. and Joe Q. 8-6: Curtis K. and Bonus 8-7: Bonus and Bonus 8-8: Bonus and Bonus

Section 8-1: Composition of Functions
Notes: [|aanotes8-1.pdf] media type="custom" key="3768323" __**Tricia L.** **& Abigail B.**__ In lesson 8-1 we learned what composition of functions were and how to use them. A composite g ò f is a function which maps x onto g(f(x)), it's domain is the set of all values in the domain f, and then also again in g. For example say you have two functions f(x)=2x+5 and g(x)=3x². When you see g o f (x) it also means f(g(x)) which tells you to work with g(x) first and then put that answer in for f. So in the functions I gave we will find g(f(x)) and then f(g(x)). In these equations x is just the place holder for another number. For the first composite of functions we first find f(x) which is g(2x+5) and then we substitute that answer in for the function g. So we get 3(2x+5)²=3(4x²+10x+25)=12x²+30x+75. Then you can also do it the opposite way which is f(g(x))=f(3x²)=2(3x²)+5=6x²+5. Then if you had a number in for the variable x you would use that and continue to simplify the composite of functions. Basically all you are doiProxy-Connection: keep-alive Cache-Control: max-age=0 roxy-Connection: keep-alive Cache-Control: max-age=0 20is substitution which we learned in a previous chapter. You always do the inside function first and then work your way out!
 * Lesson 8-1:** **Composition of Functions**

Section 8-2: Inverses of Relations
Notes: [|aanotes8-2.pdf] media type="custom" key="3775731"

In Chapter 8 Lesson 2 we learned about inverse of relation. An inverse of a relation is the relationship obtained by reversing the order of coordinates of each ordered pair in the relation. In all function the domain of g= range of f. And Range of g= domain of f=. We also learned about the inverse of a relation theorem. This states 1. A rule for g can be found by switching x and y. 2. The graph of g is the reflection of f over the line y=x. 3. The domain of g is the range of f; the range of g is the domain of f. You can figure out if the inverse is a function by the horizontal-line theorem. If it intersects the graph more than once it is not a function but if it only intersects once it is a function. - Josh H. - Andrew E.

Section 8-3: Properties of Inverse Functions
Notes: [|aanotes8-3.pdf] media type="custom" key="3788881" Chapter 8-3: In this lesson we learned the properties of inverse functions. We began the lesson by first reviewing what we saw in chapter 8-2. We saw that to find a inverse function you first have to switch the x and y. When graphing you find that a inverse function is a reflection over the line y=x. Also we know the domain and range know for inverse functions, the domain of g= range of f, the range of g= domain of f. After we reviewed what we already knew, we were given a question, what is an inverse? The answer to this question is something that "undoes" something that was already done. Inverse functions theorem states that two functions f and g are inverse functions if and only if 1.) for all x in domain of f, g • f(x)=x 2.) for all x in domain of g, f • g(x)=x To find inverse functions you first switch x and y. You then add or subtract a b value if there is one. Next you multiply or divide the value with y. To check you work you can use f(g(x)) and g(f(x)) to see if you get the value of x. We also learned the Inverse Function Notation. The Inverse Function Notation states for a function f(x), the inverse if f^-1(x), f^-1(x) is inverse. We also learned the Power Function Inverse Theorem. This says if f(x)=x^n and g(x)=x^1/n and the domains of f and g are the set of non-negative real numbers, then f and g are inverse functions. -Amanda T

In this lesson we learned more about inverse functions. As a review from the chapter 8-2 we remembered that by switching x and y we can find a inverse function and the graph of a inverse function is a reflection over the like y=x. From that lesson we also learned that the domain of g=range of f, and the range of g=the domain of f. An inverse is something that "undoes" something that was already done. Two functions f and g are inverse functions if and only if 1. for all x in the domain of f, g(f(x))=x 2. for all x in the domain of g, f(g(x))=x Switching x and y is the first step to finding inverse functions. You then add or subtract a b value (if there is one to subtract). You then either multiply or divide the value with y. If you check your work you should end up with the value of x. We also learned about the inverse function notation. This states that for a function f(x), the inverse is f^-1(x). From this lesson we also know that the power function inverse theorem says that f(x)=x^n and g(x)=x^1/n and the domains of f and g are the set of non negative real numbers, then f and g are inverse functions. -Brittani H.

Section 8-4: Radical Notation for //n//th Roots
Notes: [|aanotes8-4.pdf] media type="custom" key="3810055"

Well, in 8-4, we learned about the nth root of x. For example, if x = 4, and n = 2, this means the square root of 4 or (v'''4) which equals 2. Another example would be the 5th root of 100000 or ⁵v'''100000 which equals 10. As long as n is greater than or equal 2, this problem can be solved. Using a calculator is easy for these situations. Enter the n, click "math", press "5", then enter x and press "enter". Now if x is raised to a power, we use the root of a power theorem where if we take the nth root of x^m, the answer is x^(m/n). Example: ³v'''(100^6) = (100^6)^(1/3) = 100^(6/3) = 100^2 = 10000. Sometimes these problems appear: vvv'''(x). This is easy to figure out: (((x^1/2)^1/2)^1/2). Since the v'''x = x^1/2, just multiply the nth roots by each other. And to turn x^n into a radical, just use this the nvx (<--- the n should be associated as the 'nth root' not 'n times vx). So that was lesson 4. - Jake L.

Section 8-5: Products with Radicals
Notes: [|aanotes8-5.pdf] media type="custom" key="3816883" __**Paige H.**__ 8-5 was all about products with radicals. This was all centered around the Root of a Product Theorem, which just added onto things we already knew. Just like (xy)^m=x^m times y^m, it tells that the nth root of x times y equals the nth root of x times the nth root of y. We used this idea to multiply numbers with an nth root, such as the third root of four times the third root of 6. We then applied this idea with simplifying and writing as one radical. A radical can be simplified by breaking it up into factors that are perfect roots. The process of simplifying an nth root is simply when you factor the expression in the radical into perfect nth powers. We also added onto the idea of a mean, which is the average. To find a mean, all the numbers are added up and divided by the number of numbers you added. Adding on to this idea, a geometric mean is found by multiplying all of the values and then taking the nth root.

Section 8-6: Quotients with Radicals
Notes: [|aanotes8-6.pdf] media type="custom" key="3831019" In lesson 8-6 we talked about quotients with radicals. Radicals are numbers taken to a root and quotients are answers to division problems. We learned how to rewrite a fraction without a radical in the denominator. To solve this multiply the numerator and denominator by the radical, and this will not change the value of the fraction. The process of writing a fraction in an equivalent form without irrational numbers in the denominator is called rationalizing the denominator. We do this to recognize equivalent expressions. To rationalize the denominator of a fraction whose denominator is a square root, multiply both numerator and denominator by the square root. Now when the fraction has a radical in the denominator is added to another term we use a technique thats similar to the one we used to divide complex numbers. -Andrew E.

This lesson is all about rationalizing the denominator of a fraction when there is a radical in it. To rationalize a fraction with a radical in the denominator we must simply multiply both the top and bottom parts of the fraction by whatever the radical is we are trying to get rid of. What this does is rewrite the radical in a different form that serves the same purpose. From here, just simplify. A radical times a radical will equal whatever is inside the radical. -Curt K.

Section 8-7: Powers and Roots of Negative Numbers
Notes: [|aanotes8-7.pdf] media type="custom" key="3841337" In Section 8-7: Powers and Roots of Negative Numbers, we learned about rules concerning positive and negative roots and the numbers being rooted. We remembered that with (-x)^n, the answer is positive when n is even and negative when n is odd. We also learned that we can't combine the square root of x and the square root of y when they're negative. For example, the square root of -6 times the square root of -10 will not equal the square root of sixty. The square roots of negative numbers will have i in them, which won't equal the square root of sixty. We found that we can take the nth roots of negative numbers when n is odd. We can't do this when n is even. -Emma R

In lesson 8-7, we dealt with applying the same properties we learned for positive bases but instead used negative bases. We figured out that, when dealing with negative numbers and the (-x)^n, the domain of the equation is positive when n is even and negative when n is odd. This is due to the fact that the square root of a negative number is imaginery. Its important to remember that you have to be careful what you simplify when a base is negative. Make sure you keep track of your negative signs and dont skip steps or its likely you will make a mistake. -Trisha T.

Section 8-8: Solving Equations with Radicals
Notes: [|aanotes8-8.pdf] media type="custom" key="3848813" In 8-8, we started the lesson with extraneous solutions, which is where in a radical problem the final answer does not match the original answer when checked. Example: 6 - v'''(x-2) = 88. Subtract 6, -v'''(x-2) = 82. Square both sides, -(x-2) = 6724. Distribute the negative on (x-2), -x + 2 = 6724. Subtract 2, -x = 6722. Divide by -1, x = -6722. Now check your answer by substituting -6722 into 6 - v'''(x-2) = 88. Your final answer is not equal to 88. We also dealt with the distance formula which assisted in finding two points on a vertical line where there was only one point on the grid (not on the line). Sometimes when solving equations with radicals, you end up with two answers: v'''(4x^2) = 10. Simplify, ±2x = 10. The reason the ± is there is because 2x can be positive or negative when squaring them. When solving this, you divide both sides by 2 and -2. The answers are 5, -5. In some problems, if x is said to be greater than 2, than you would choose the positive 5, not negative 5. And that is lesson 8. - Jacob L.

__**Paige Heeter**__ Lesson 8 was called Solving Equations with Radicals. We first saw the idea that we can use ideas from previous things we learned and apply it to doing the same thing radicals. The main idea in solving these types of problems is to try to get the radical by itself and then put it to whatever root the radical is being rooted by. Then we learned about extraneous solutions. This is a solution that is found when solving an equation with a radical which is not a solution to the original problem. After seeing many examples of how to solve equations with radicals, we learned how to do problems that ask you to find two points on a certain line that is x units away from a certain point. To solve this type of equation with radical we used the Distance Formula. This is d=the square root of (x1-x2)^2 + (y1-y2)^2. We then got our homework assignment which gave us much help and practice with this lesson.

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