Task 1)

Part 1) Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses.

f(x)=

x+ab

g(x)=cx−dPart 2. Show your work to prove that the inverse of f(x) is g(x).

Part 3. Show your work to evaluate g(f(x)).

Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include five values for each function. Graph the line y = x on the same graph.

Task 2

Part 1. 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 first equation has an extraneous solution and the second does not.

Part 1. Using the two functions listed below, insert numbers in place of the letters a, b, c, and d so that f(x) and g(x) are inverses.

f(x)=x+ab

g(x)=cx−d

Answer:

Let a=2 , b=7 , c=1 , d=14

f(x)=x+(2)(7)

f(x)=x+14

g(x)=(1)x−(14)

g(x)=x-14

 

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

Proof:

y=x+14

interchange x and y, and solve y in terms of x.

x=y+14

x-14=y

y=x-14

 

Part 3. Show your work to evaluate g(f(x)).

Answer:

We just need to plug in the value of f(x) to x of g(x)

g(f(x))= (x+14)-14

g(f(x))= (x+14)-14

g(f(x))= x

Part 4. Graph your two functions on a coordinate plane. Include a table of values for each function. Include five values for each function. Graph the line y = x on the same graph.

The values are given below:

Task 2

Part 1. 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.

 

****Extraneus Solution- Extraneus Solution is a solution that is really not part of the solution.

example:

let a=6 , b=2 , c=3 ,d=4

6√x+2+3=4

6√x+5=4

 

****without extraneus soltuion

let a=1 ,b=2 ,c=3 , d=4

1√x+2+3=4

√x+5=4

 

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

Solve:

6√x+5=4

√x+5=4/6 square both sides

(√x+5)²=(4/6)²

x+5=16/36

x+5=4/9

x=(4/9)-5

x=(4-45)/9

x=-41/9

 

****show and check that your solution is extraneous:

plug in -41/9 in 6√x+5=4

6√(-41/9)+5=4

6√(-41+45)/9=4

6√4=4

6(2)=4

12=4

Since 12 is not equal to 4 then -41/9 is extraneous solution.

 

Solve:

√x+5=4 square both sides

(√x+5)²=4²

x+5=16

x=11

 

Check:

√x+5=4

√11+5=4

√16=4

4=4

 

since 4=4 then √x+5=4 has no extraneus solution

 

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

Answer:

Notice that the only difference between the 2 equation is the variable a, :

6√x+5=4 a=6

√x+5=4 a=1

the reason why the first one has extraneus solution is because d/a=4/6=2/3 is not a whole number and the reason why the second one has no extraneus solution is because d/a=4/1 =4 is a whole number.

Please see the attached file for the complete solution