Question

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+a
b

g(x)=cx−d

Part 2. I need to show my work to prove that the inverse of f(x) is g(x).

Part 3. I need to show work to evaluate g(f(x)).

Part 4. I need to graph 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.

Answer 1

@ryanvarghese12790 Hey man! I was told to come to you for help by tester97, if you could help me that would be awesome.

Answer 2

Okay, how about we do f(x) = 3x -5 ?

well, to get g(x) you just find the inverse of f(x) like this:

f(x) = 3x -5
Now change f(x) for
y= 3x -5
Switch the x and y
x=3y -5
Now isolate the y by adding five to both sides
x + 5 =3y
Now divide both sides by 3
\[ \frac{ x +5 = 3y }{ 3 }\]\[y = \frac{ x +5 }{ 3 }\]

For part one, you just write the functions, and then for part 2, you show the work that I showed to find the inverse.

Answer 3

Wow! Thanks man :-) I have a few other things though :
\

Answer 4

@ryanvarghese12790 If you can help me with them please?

Answer 5

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.

Answer 6

if you take the sample they gave you,
a√(x+b)+c=d*x
like i said, I replaced d with d*x because I believe you mistyped it.

rearranging, you get
a√(x+b)+c -d*x = 0

Remember, that a, b, c and d are all just numbers. The difference between a,b,c,d and x is that when we actually try to solve this, we will put in actual numbers (like 1 or 3.2 or 9/2) for a,b,c,d but we won't for x. After we have replaced a,b,c,d with numbers, we will then try to find the value for x that makes
a√(x+b)+c -d*x = 0

In the calculator I linked, it plots
y = a√(x+b)+c -d*x
and there are sliders for a,b,c and d so you can pick what number thy are.

but as you saw above, the actual solutions for x are when
a√(x+b)+c -d*x = 0
by comparing that to
a√(x+b)+c -d*x = y

we can see that the solutions on the graph are when y = 0

so if you open the graph and move around the sliders for a,b,c and d, you can see that sometimes the curve intersects the x axis in one place, and sometimes it does in 2 places. When it intersects in one place, that means you have 1 solution.
to find a radical expression that has one solution, mess with the sliders until the curve only intersects the x axis at 1 point. Then, write out
a√(x+b)+c -d*x = 0
take your values for a, b, c and d that were in the calculator sliders and put them into your equation. So say in the calculator , a = 5, b = 2, c = 1 and d = 1. In that case you would write
5√(x+2)+1 -1*x = 0

Then solve that equation for x (square everything and use the quadratic formula)
After you have your 2 x values, plug them both back into
5√(x+2)+1 -1*x = 0
If the value for x makes the left side not equal 0, it is an erroneous solution and you can throw it away. If it does make it equal 0, it is a good solution.

To find a radical expression with 2 solutions, first change the constants in the calculator so that the curve intersects in 2 places, then repeat the stuff from above.

Answer 7

@ryanvarghese12790 ilysm

Answer 8

does that mean I love you so much?

Answer 9

@ryanvarghese12790 yup

Answer 10

And I have one more but idk if you'll like to help me