Unit+5+Virtual+Notebook+Christmas+Break+Assignment

__**Part 1**__ Given the function g(x) = x^2 - 7 when x<=0 1.) Find the inverse algebraically and explain your step-by-step process verbally. To find the inverse, first we must flip the x and y in the equation, making it x = y^2 - 7. Then, we must solve for y. Add 7 to both sides, then find the square root, so the inverse of g(x) is y = sqrt(x+7)

2.)The graph of g(x) is labeled below. Explain how you can find at least 4 coordinate points that represent the inverse function. Write the coordinate points you found. You can find 4 points that represent the inverse function by first finding 4 points of the original function. Four points of g(x) are (0,-7), (-1,-6) , (-2,-3), and (-3,2). To find the inverse, we simply flip the x and y values, so 4 points that represent the inverse of g(x) are (-7,0) , (-6,-1) , (-3,-2) , (2,-3).

3.) Explain why g(x) was given the domain x<=0 rather than sketching the entire function when finding the inverse graphically. For both positive and negative values of x, the y value would be the same thing, since g(x) is a quadratic function. However, graphically, the graph of the inverse would only show the positive values of y.

4.) Given the domain of a function h(x) is [-3, infinity) and the range is [0, infinity), find the domain and range of the inverse function. Explain how you arrived at your answer. The domain of the inverse function is [0, infinity) while the range is [-3, infinity); the inverse function would merely have the intervals flipped for domain and range.

1.) Explain the step by step process of graphing the function g(x) = 1 - 2log3(x+4) (this is a log base 3, the 3 should not be distributed through the parenthesis) without using the graphing calculator. Find three point of the parent function and the three corresponding points of g(x). List the asymptote(s), domain, and range. To graph, first find the points of the parent function, 3^x. 3 points would be (0,1), (1,3), and (2,9). Next, to find the inverse, log3(x), flip the x and y values of the points of the parent function. We must also shift 4 units to the left, multiply the y values by -2, and go up one unit. Thus, 3 points of g(x) are (-3,1) , (-1,-1), and (5,-3). The asymptote of the function is x = -4, the domain is (-4, infinity) and the range is all real numbers.
 * __Part 2__**

2.) Explain how you can find the domain and range of any logarithmic function without looking at the graph or using a graphing calculator. The domain of a logarithmic function would be the value the x is added/subtracted by up to infinity. The range is all real numbers.


 * __Unit 5 Lesson 5 __**

In unit 5 lesson 5 we evaluated logarithmic and exponential expressions without using a calculator. Evaluate the following expression. Since I cannot control whether you use a calculator at home you must write your steps out verbally so I know you understand the process.

a. It is equal to ln1 - lne^(2/5), which is equal to 0-2/5, which equals -(2/5)

b. Using the power rule: equal to 1/4log3(27); equal to 1/4(3) = .75

c. Power rule: 3log2(16), which equals 3(4) = 12

d. lne cancels out, so it equals 2-3, which results in -1

e. It is equal to log4(1) - log4(16), which equals 0 - 2, which is -2.

f. This equals log5(2) - log5(250); however, 2/250 can be simplified to 1/125, so we can change the expression to log5(1) - log5(125), which is equal to 0 - 3, which is -3.

In Unit 5 Lesson 6 We learned about rewriting logarithmic expressions by expanding to have multiple logarithms and condensing to have a single logarithm.
 * __Unit 5 Lesson 6__**

I want you to prove algebraically, why the following statements are true using properties of logarithms.

a.Multiply both parts of the fraction (1/250) by .008; this will make the left side of the equation log5(.008/2), which is equal to log5(.008) - log5(2), which is equal to -3 - log5(2).

b. Using the product and power rules, 3ln2 = ln2^3, which is ln8, and that + ln3 would be equal to ln(8*3), which is ln24; add the negative sign in front and it is -ln24.

After break we will be learning how to solve various logarithmic equations and how to do applications algebraically; however if you understand the concept of exponential applications your should be able to solve and analyze a logarithmic application.
 * __Unit 5 Lesson 7__**

a. Plugging in 0 for t should find us the initial amount; the result of f(0) is 90 units of initial drug administered.

b. The amount of drug is 66 mg after .5865 hours. This answer was obtained by plotting f(t) and y = 66 on the same graph and then finding the intersection.