Question

3 worded answers help Medal to helper

Answer 1

Find the inverse.

Answer 2

Switch and solve.

Answer 3

I dont know how please can you help me step by step?

Answer 4

That's the best I can do in 3 words. lol

Answer 5

Okay nvm

Answer 6

You have:

\(\Large f(x)=\dfrac{4x-3}{2}\)

To find the inverse, you want the function that "swaps" the domain and range. Usually we do that with the switch and solve method.

First, replace f(x) with y, so you have y=.... notation:

\(\Large y=\dfrac{4x-3}{2}\)

Now "switch" the x and the y - replace y with x, and x with y:

\(\Large x=\dfrac{4y-3}{2}\)

Answer 7

Now, SOLVE for y. That will be your inverse function.

Answer 8

Okay that helps thanks

Answer 9

The original function has "input" of # of weeks in course, and "output" of how many assignments have been completed.

The inverse switches the domain and range, so inverse has "input" how many assignments have been completed, and "output" number of weeks in course.

Hence, you can evaluate the inverse for x=30, to find how many week it will take to have completed 30 assignments.

Answer 10

but how do you solve for x thats my question

Answer 11

how do you solve 4y-3/2

Answer 12

I think the wording of the question, "solve for x=30" is odd.... what it really means is, find the value of the inverse at x=30. E.g., find:

\(\Large f^{-1}(30)\).

ONce you have the inverse function, just plug in x=30 and find the function value.

It has exactly the same result as if you just set the ORIGINAL function = 30 (e.g., let y=30) and then solve for x. Because remember, the original and the inverse functions just have reverse domain and range.

Answer 13

Do you mean, how do you solve THIS for y?

\(\Large x=\dfrac{4y-3}{2}\)

Answer 14

yes

Answer 15

Just isolate y. Just like you'd solve any equation for a specified variable.

Get the y alone, so you have y={some stuff that involves x}

You'll want to start by multiplying both sides by 2. After that it's all downhill. :)

Answer 16

okay thanks

Answer 17

\[\Large 2\cdot x=\left(\dfrac{4y-3}{2}\right)\cdot 2\]

Answer 18

you're welcome.

Answer 19

thanks alot
so it would be 2x=8y+6?

Answer 20

cause the 2s cancel out on the right side of the equation
then u plug 30 to x?

Answer 21

no, the 2's cancel on right hand side.

\(\Large 2\cdot x=\left(\dfrac{4y-3}{\cancel 2}\right)\cdot \cancel 2\)

Answer 22

If you canceled the 2's on the RHS, then why did you also multiply the 2 through the 4y-3?

Answer 23

And no, you don't plug in x=30 yet... you need to finish finding the inverse. Solve for y!

Answer 24

ohh okay lol sorry
so its 2x=4y-3

Answer 25

add 3 to both sides?

Answer 26

yes, what next?

Isolate the y. That's your goal.

Right, add 3.

Answer 27

so its 5x=4y?

Answer 28

then you divide?

Answer 29

nope. you added 3 to both sides, so should have gotten:

2x+3=4y

You CANT COMBINE the 2x and the 3. they are NOT like terms.

Answer 30

ohh okay

Answer 31

NOW divide.

Answer 32

dive 4?

Answer 33

to 2x+3

Answer 34

right

Answer 35

how do you dive that by 4

Answer 36

Just put it in a numerator, over 4 in the den'r.

\(\Large y=\dfrac{2x+3}{4}\)

Answer 37

Remember, dividing by 4 is really just multiplying by 1/4.

Answer 38

so multiply both sides by 4?

Answer 39

Whoaah.... why would you multiply both sides by 4?? We just DIVIDED by 4. Multiplying by 4 would just "undo" that.

Answer 40

THAT is your inverse function:

\(\Large f^{-1}(x)=\dfrac{2x+3}{4}\)

You should know that notation: \(f^{-1}(x)\) means the "inverse function of f(x)".

Answer 41

okay im confused so you divide by 4 but how do I do that

Answer 42

OK, let's back up.

You were to this point:

\(\Large 4y=2x+3\)

ALL that's left, to solve this equation for y (which really just means to "get the y all alone on one side of the = sign"), is to divide both sides of the equation by 4, right?

When we do that on the left, we have y all alone. So we just have to do it on the right, also.

Answer 43

I have to go thanks for the help :)

Answer 44

\(\Large \dfrac{1}{4}\cdot4y=(2x+3)\cdot\dfrac{1}{4}\)

\(\Large y=\dfrac{2x+3}{4}\)

OK, bye.