Question

find f^-1 and verify that (fof^-1)(x)=(f^-1of)(x)=x

f(x)=2x+3

Answer 1

how would you solve
\[2x+3=7\]?

Answer 2

1) subtract 3
2) divide by 2

that is what the inverse does, subtract 3, divide by 2, i.e.
\[f^{-1}(x)=\frac{x-3}{2}\]

Answer 3

if you like you can write
\[x=2y+3\] and solve for \(y\)
you get
\[x=2y+3\\x-3=2y\\
\frac{x-3}{2}=y\] i.e.
\[f^{-1}(x)=\frac{x-3}{2}\]

Answer 4

ok thank you. when the problem asks to find (fof^-1)(x)=(f^-1of)(x)=x, I know I am supposed to plug in the inverse in every x of the original equation and vice versa, but I'm not exactly sure how both will end up equaling just x.

Answer 5

bunch 'o algebra

Answer 6

For the verification part, you have
\[f(x)=2x+3\\
f^{-1}(x)=\frac{x-3}{2}\]
So,
\[\begin{align*}\left(f\circ f^{-1}\right)(x)&=f\left(f^{-1}(x)\right)\\
&=f\left(\frac{x-3}{2}\right)\\
&=2\left(\frac{x-3}{2}\right)+3\\
&=x-3+3\\
&=x
\end{align*}\]
The procedure is identical for determining \(\left(f^{-1}\circ f\right)(x)=x\).

Answer 7

\[f(f^{-1}(x))=2(\frac{x-3}{2})+3\]

Answer 8

if your inverse is correct you will end up with a raft of cancellation, giving you nothing but \(x\) at the end

Answer 9

alright thank you both so much. I was just really confused with the way the problem was written.