Question

Given f(x) =5x+1/2solve for f^1(8)

3

7

10

20

Answer 1

Is that \(f^{-1}(8)\)

Answer 2

yea

Answer 3

the equation is\[f ^{-1}(8)=5x+1\div2\]

Answer 4

Ugh! Austin you always beat me to them XD

Answer 5

\(\displaystyle f^{-1}(x)=5x+\frac{1}{2}\)

Or

\(\displaystyle f^{-1}(x)=\frac{5x+1}{2}\)

Answer 6

the second one

Answer 7

wait it not negetive power it is positive one\[f^{1}\]

Answer 8

Actually, that would be \(\displaystyle f(x)=\frac{5x+1}{2}\)

We need to find \(f^{-1}(x)\)

How we would do this is as follows.

Set f(x)=y

so now we have,

\(\displaystyle y=\frac{5x+1}{2}\)

Now we need to solve for x.

\(\displaystyle y=\frac{5x+1}{2}\)

\(2y=5x+1\)

\(2y-1=5x\)

\(\displaystyle x=\frac{2y-1}{5}\)

Now we switch the x and y.


\(\displaystyle y=\frac{2x-1}{5}\)

\(\displaystyle f^{-1}(x)=\frac{2x-1}{5}\)

Now we need to find \(f^{-1}(8)\)

Which all we would do is simply replace all instances of x with 8 in the function. Do you think you can do that?

Answer 9

yea but change the f^-1 to f^1 it glitch when i typed it

Answer 10

That makes absolutely no sense... \(f^{1}(x)=f(x)\)....

Answer 11

idk either but tht is how it is on my test

Answer 12

wait.... this is a test?

Answer 13

it a redo.i did the question but i got it wrong the first time.it confused me so she said i could ask for help on tht question

Answer 14

Ok then... well... just plug in 8 for all instances of x in f(x)

Answer 15

ok i did and i got [40+1 \div2\]

Answer 16

\(\displaystyle f(8)=\frac{5(8)+1}{2}=\frac{41}{2}\)

That should be your answer with the weird notation and all.

Answer 17

oh i forgot to add the one.ok then would i divide?

Answer 18

You can leave it like that, or you can make it an improper fraction. But that should be acceptable as is.

Answer 19

ok thank you

Answer 20

No problem!