Find a formula for the inverse of the function. f(x)=(4x-1)/(2x+3) y=ln(x+3) y=x^2 - x y=(1+e^-x)/(1+e^-x)

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Find a formula for the inverse of the function. f(x)=(4x-1)/(2x+3) y=ln(x+3) y=x^2 - x y=(1+e^-x)/(1+e^-x)

Calculus1
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if the function is one-to-one, you just solve the equation for x
\[y=\frac{ax+b}{cx+d} \\ \text{ multiply both sides by } cx+d \\ (cx+d)y=ax+b \\ \text{ distribute on the left } \\ cx y +dy=ax+b \\ \text{ now remember you want to solve for } x \\ \text{ get all of your terms with } x \text{ on one side } \\ \\ \text{ so subtract } ax \text{ and } dy \text{ on both sides } \\ cxy-ax=b-dy \\ \text{ now this allows you to do the following so you can solve for } x \\ x(cy-a)=b-dy \\ \text{ now divide both sides by } (cy-a)\]
Oh, ok. Thank you! :)

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after solving for x,interchange x any y
y=(1+e^-x)/(1+e^-x) How about the very last one? The negative exponent is confusing me.
if the negative exponent confused you can multiply top and bottom by e^x
\[y=\frac{1+e^{-x}}{1+e^{-x}} \cdot \frac{e^{x}}{e^{x}} \\ y=\frac{e^{x}+1}{e^{x}+1}\]
wait a minute
lol (1+exp(-x))/(1+exp(-x)) is 1
is there a type-o?
\[y=\frac{1+e^{-x}}{1+e^{-x}}=1 \]
and f(x)=1 is not one-to-one
I think the typo is that the numerator is (1-e^-x) while the denominator is (1+e^-x). But would that make that much of a difference?
Oh wait, it would. You can't simplify it to be 1.
yes! now the top is different from the bottom the function is not 1 anymore now verify the function is one-to-one and if it is solve for e^x and then solve for x and if is not you are done because no inverse function exist
I said solve for e^x because you can still multiply top and bottom by e^x giving you \[f(x)=\frac{e^{x}-1}{e^{x}+1}\]
Ah I see how to solve it now. Thanks again!
are you looking for inverse relation or function?
Inverse function
so did you verify f(x)=(1-exp(-x))/(1+exp(-x)) is one-to-one? you can use a graphing calculator and see if it passes the horizontal line test?
Oh ok. So I plugged it into the graphing calculator and the function doesn't pass he horizontal line test.
the*
Actually, it does. Nevermind, it looked like it didn't at first.
ok so that means you have to actually solve that equation above for e^x then x
Alright, will do.

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