AmTran_Bus
  • AmTran_Bus
If f(x)=2x-1/3 and f^-1 is the inverse of F, what is the value of f^-1(3)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
about 1.67
anonymous
  • anonymous
because f^-1(x)=(1/2)x+1/6
jdoe0001
  • jdoe0001
well, what's the \(f^{-1}(x) \ of \ 2x-\frac{1}{3} ? \)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

AmTran_Bus
  • AmTran_Bus
How do you get 1/3?
anonymous
  • anonymous
2^-2(3)-1/3 2^-6-1/3 do you want it as a fraction if so this is it unless you want a decimal
jdoe0001
  • jdoe0001
1/3? do you know how to find the \(f^{-1}?\)
AmTran_Bus
  • AmTran_Bus
@jdoe0001 , I need refreshing on the whole thing.
AmTran_Bus
  • AmTran_Bus
and I need it in whole number form. I have the answer, I just need to know how to get it. The answer is 5.
jdoe0001
  • jdoe0001
if the function is f(x) = y y = ax+b then the inverse is obtained by doing a switcharoo on the variables position, so it becomes x = ay+b then you solve for "y" and is x = ay+b => x-b = ay => (x-b)/a = y so the inverse for y= ax+b would be \(\large {y = \frac{x-b}{a}}\)
anonymous
  • anonymous
\[\frac{ 1 }{2 }x+\frac{ 1 }{ 6 } = f ^{-1}(x)\]
anonymous
  • anonymous
\[which is = \frac{ x + \frac{ 1 }{ 3 } }{ 2 }\]
AmTran_Bus
  • AmTran_Bus
I'm at this point. x=2y-1/3
anonymous
  • anonymous
ok good, simply add the 1/3, than divide the 2
anonymous
  • anonymous
so you will have (x+1/3)/2=f^-1(x)
AmTran_Bus
  • AmTran_Bus
\[x=\frac{ 2y-1 }{ 3 }\] Where do you get subtract 1/3?
AmTran_Bus
  • AmTran_Bus
I see the neg1, but you dont put the 2y over the 3 also?
anonymous
  • anonymous
wait ooooo, thats different
jdoe0001
  • jdoe0001
$$ \color{blue}{y} = 2\color{red}{x}-\frac{1}{3}\\ \text{doing a switcharoo}\\ \color{red}{x} = 2\color{blue}{y}-\frac{1}{3}\\ \text{solving for "y"}\\ \cfrac{\color{red}{x}-\frac{1}{3}}{2}=\color{blue}{y} = f^{-1}(x) $$
anonymous
  • anonymous
no jdoe look, he put it differntly now
jdoe0001
  • jdoe0001
wait... a sec
jdoe0001
  • jdoe0001
$$ \color{blue}{y} = 2\color{red}{x}-\frac{1}{3}\\ \text{doing a switcharoo}\\ \color{red}{x} = 2\color{blue}{y}-\frac{1}{3}\\ \text{solving for "y"}\\ \cfrac{\color{red}{x}+\frac{1}{3}}{2}=\color{blue}{y} = f^{-1}(x) $$
AmTran_Bus
  • AmTran_Bus
Here is the ORIGINAL problem: \[f(x)=\frac{ 2x-1 }{ 3}\] FInd \[f ^{-1}(3)\]
anonymous
  • anonymous
\[\frac{ 3x+1 }{ 2 } = f^1(x)\]
jdoe0001
  • jdoe0001
hmm, that looks different some :/
anonymous
  • anonymous
f^-1(3)=5
anonymous
  • anonymous
no wonder we did it wrong, you gave it to us wrong.
AmTran_Bus
  • AmTran_Bus
That is real nice. :(
anonymous
  • anonymous
so yes the answer is 5.
AmTran_Bus
  • AmTran_Bus
Can you work it out step by step? (patience please)
anonymous
  • anonymous
ok
jdoe0001
  • jdoe0001
$$ y=\frac{ 2x-1 }{ 3}\\ \text{doing a switcharoo}\\ x=\frac{ 2y-1 }{ 3}\\\\ \text{solving for "y"}\\ $$
jdoe0001
  • jdoe0001
whatever "y" gives you, is your \(f^{-1}(x)\)
AmTran_Bus
  • AmTran_Bus
Thanks, got it!

Looking for something else?

Not the answer you are looking for? Search for more explanations.