Task 1:
Part 1: Using the two functions listed below, insert numbers in place for the letters A, B, C, D so that f(x) and g(x) are inverses.
f(x)= x + a
g(x)= cx - d

Part 2: Show your work to prove that the inverse of f(x) is g(x).

Part 3: Show work to evaluate g(f(x))

Part 4: Graph you're two functions on a coordinate plane, include a table of values for each function. Include 5 values for each function.graph the line y=x on the same graph.

Task 2:
Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.
a√x + b + c = d

Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

Part 2: Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.

Part 3: Explain why the equation has an extraneous solution and the second one does not.

Task 3: Create a scenario for an arithmetic sequence. For example, Jasmine practices the piano for _______ minutes on Monday. Every day she __________ her practice time by _____________. If she continues this pattern, how many minutes will practice on the 7th day? Be sure to fill in the blank's with the words that will create an arithmetic sequence. Use your scenario to write the functionfor the 7th term in your sequence using sequence notation.

Part 2: Create a scenario for geometric sequence. For example, Anthony go to the gym for _________
minutes on Monday. Every day he __________ his gym time by _________. If you continue this pattern, how many minutes will he spend at the gym on the 5th day. Be sure to fill in the blank's with the words that will create an geometric sequence.
Use your scenario to write the functionfor the 5th term in your sequence using sequence notation.

Part 3: Use your scenario from part 2 to write a question. That will lead to using the geometric series formula. Use the formula to solve for Sn in your scenario.
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1 Answer


Part 1

Let y=f(x)=(x+a)/b; x=g(y)=by-a=cy-d, so c=b and d=a.

Let d=a=2 and c=b=3: f(x)=(x+2)/3 and g(x)=3y-2.

Part 2

y=f(x)=(x+2)/3; 3y=x+2; x=3y-2=g(y) so g(x)=3x-2, the inverse.

Part 3

g(f(x))=3f(x)-2=3(x+2)/3-2=x+2-2=x, so g(f(x))=x showing that f(x) and g(x) are mutual inverses.

x= -2 -1 0 1 2
f(x) 0 1/3 2/3 1 4/3
g(x) -8 -5 -2 1 4


Part 4


Part 1

(a) Let a=1, b=2, c=3, d=6: √x+2+3=6; √x-6=-5; (b) Let a=-1, b=2, c=3, d=4

(a) Multiply both sides by √x+6: x-36=-5(√x+6)=-5√x-30; x-36+30=-5√x; x-6=-5√x.

Square both sides: x^2-12x+36=25x; x^2-37x+36=0=(x-36)(x-1).

Part 2

(a) So x=36 or 1 (apparently). Substitute x=36 in the original equation: 6+2+3=6, 11=6 is not true. So x=36 is an extraneous solution; substitute x=1: 1+2+3=6 is true, so x=1 is the actual solution.

(b) -√x+2+3=4; -√x=-1; multiply both sides by -1: √x=1; square both sides: x=1. Substitute x=1 in the original equation: -1+2+3=4 is true. So x=1 is the solution.

Part 3

For (a) the act of squaring both sides resulted in creating an extraneous solution ((-5)^2=25=5^2). For (b) the solution was simpler and there was no ambiguity.


Part 1

John read 2 chapters of a book on Sunday. Each day of the week he reads one chapter more than the day before. How many chapters did he read on Saturday? How many chapters had he read altogether by the end of Saturday? 

On Sunday he read 2 chapters; by Monday he'd read 3 more chapters; by Tuesday, 4 more chapters; by the following Saturday he'd read 8 more chapters. 

a=2, d=1. The sequence is a, a+d, a+2d, ... For n days the term is a+d(n-1). (The sum is na+nd(n-1)/2.) Plug in a and d, and n=7: 2+6=8 for the 7th term. So he reads 8 chapters on Saturday. (The sum is 2*7+7*6/2=14+21=35 chapters.)

Part 2

On day 1 a bacterial culture contained 10000 bacteria. Each day the bacteria multiplied to increase their numbers by 10%. How many bacteria will there be on day 5?

a=10000 and r=1.1. The series is 10000, 11000, 12100, 13310, 14641.

The nth term is ar^(n-1)=10000*1.1^4.

Part 3

Each day the bacteria are given 1g of nutrient per 10000 bacteria. How much nutrient will they need for n days?

Sn is the total quantity of nutrient they will need after n days=1+1.1+1.1^2+...+1.1^(n-1)=(1.1^n-1)/(1.1-1)=10(1.1^n-1) grams.

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