what is f^-1(4) when f(x)=(3x-4)/5?
@ranga help?
@ParthKohli help?
You need to find the inverse of f(x). To do that follow these steps: 1. Replace f(x) with y. 2. Interchange x and y. That is, put x in the place of y and y in the place of x. 3. Solve for y. This will be the inverse. 4. replace y with f^-1. 5. Find f^-1(4) by putting x = 4 in the inverse function.
so f^-1(4) would = -4?
I m getting a different answer. If you show me the steps I can try to spot any mistakes in the steps.
i dont know how much is f^-1(4)
First you have to find the inverse of the function. I gave you the steps above. What is the inverse function you are getting? f^-1(x) = ?
3y-4/5?
What you have is step #2. x = (3y - 4) / 5. In step#3, solve for y. That is y = ?
how would i solve for y ?
x = (3y - 4) / 5 To solve for y, isolate y. multiply both sides by 5. Add 4 to both sides. divide both sides by 3.
i cant do it i nee to see step by step :/
x = (3y - 4) / 5 multiply both sides by 5. What do you get?
so it would be x=15y-20?
@ranga
When you multiply the left hand side by 5, x will become 5x. When you multiply the right hand side by 5, it will cancel out the 5 in the denominator and you will be left with just (3y - 4)
ooh alright got it now whats next? replace y with f^-1 so it'll be 5x=3f^-1-4?
Not yet. You have to solve for y first. x = (3y - 4) / 5 multiply both sides by 5: 5x = 3y - 4 add 4 to both sides: 5x + 4 = 3y divide both sides by 3 (5x+4)/3 = y or y = (5x + 4) / 3 Now do step 4 and replace y with f^-1(x) f^-1(x) = (5x + 4) / 3 f^-1(4) = ?
-4 @ranga
Then just plug in -4 into the x?
Not -4, plug in +4.
What do you get when you put x = 4 in (5x + 4) / 3 ?
i get 8
Yes.
so thats the answer?
Yes.
thank you @ranga you're the best!
You are welcome. Thanks.
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