Unit+5+Virtual+Notebook

__**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. **g(x)^-1 = square root of (x + 7)** **What i did was switch the x and y from the original equation and solve the y from there, which should result into the inverse.** 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. **All that is needed to be done is to take the original points and switch the x and y coordinates.** **4 Coordinate Points- (-4,9) (-3,2) (-2,-3) (-1,-6)** **Inverse from the Original, respectively- (9,-4) (2,-3) (-3,-2) (-6,-1)** 3.) Explain why g(x) was given the domain x<=0 rather than sketching the entire function when finding the inverse graphically. **This is because when the inverse is taken, the domain of the inverse would turn to (all real numbers) and the range would have x<=0** 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 answer is simple, all that is needed to be done is switch the domain and the range, yet the answers are the same. In this case, the domain and range of the inverse would be D= [0, infinity) and the R= [-3, infinity) of h(x)^-1** 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. __**Parent function 3^x**__ **x= -1 0 1** **y= 1/3 1 3**
 * __Unit 5 Lesson 4__**
 * [[image:rogowiczprecalc/Unit_5_Lesson_4_Journal.png width="577" height="331" caption="Unit_5_Lesson_4_Journal.png"]] ||
 * __Part 2__**

**log3** **x= 1/3 1 3** **y= -1 0 1**

**4 units left** **x= -3 2/3 -3 -1** **y= -1 0 1**

**Stretch by -2** **x= -3 2/3 -3 -1** **y= 2 0 -2**

**1 unit up** **x= -3 2/3 -3 -1** **y= 3 1 3**

**VA= x= -4** **D= (-4, inf.)** **R= (all real numbers)**

2.) Explain how you can find the domain and range of any logarithmic function without looking at the graph or using a graphing calculator. **Remember that there are no HA's, so that would mean that the range will always be all real numbers. as for the domain, it could be either (0, inf) if there is no 's, or if there is only x. however, if there is a number in the parenthesis with x, there just take the value in that  and set it equal to or greater than 0. (this is confusing, so i will use an example- if you have f(x)= log3(x-3), take the (x-3), set it equal to or greater than 0, then solve for x. this will be x>= 3, therefore that domain is D= (3, inf) )**

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) Set equal to x, change the denominator of fifth root e^2-->e^2/5. then cross out the ln and e, and x = 1/(2/5), or 2.5** **B) set equal to x, then change to 3^x because you want to cross out the log3. in doing that, you have fourth root of 27, therefore the x of 3^x would be 3/4, because 3^(3/4) is 4th root of 27.** **C) set equal to x, then change it to 2^x to cross out the log2. what is left is the 16^3. know that 2^x = 16^3, and that 2^4 equals 16, and multiply that by 3 because of the 3 on the 16. therefore x = 12.** **D) set equal to x, and then cross out the lne's, therefore, you have 2 -3, and x = -1.** **E) set equal to x, then change to 4^x because you want to cross out the long4, then x = -2 becuase 4^-2 is the same as 1/16** **F) set equal to x, take 2/250-->1/125, then do 5^x = 1/125. x= -3**
 * __Unit 5 Lesson 5 __**


 * [[image:rogowiczprecalc/Unit_5_Lesson_5_Journal.png caption="Unit_5_Lesson_5_Journal.png"]] ||
 * Unit_5_Lesson_5_Journal.png ||

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) each side equals -3.43, because i merely solved each side. on the left hand side, i did log(1/250) divided by long(5)= -3.43, and on the right hand side, log(2)/log(5) - 3, which would also equal -3.43. Therefore, the equation is correct.** **B) i took the -(3ln2 + ln3)--> -(ln8 + ln3)--> -(ln24), so both sides would be the same**
 * [[image:rogowiczprecalc/Unit_5_Lesson_6_Journal.png caption="Unit_5_Lesson_6_Journal.png"]] ||
 * Unit_5_Lesson_6_Journal.png ||

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. **A) the initial amount 90 mg** **B) 66 = 90 - 52ln(1 + t)** **solve the t from there and you get, and you get t = .5865, which was around 35 minutes.**
 * __Unit 5 Lesson 7__**
 * [[image:rogowiczprecalc/Unit_5_Lesson_7_Journal.png caption="Unit_5_Lesson_7_Journal.png"]] ||
 * Unit_5_Lesson_7_Journal.png ||