sum1 here help!!!

[sonya]

ok....since no1s helpin in the hmwk section thot id take a shot here..

if f(x)= (2e^x -8)/(10e^x + 9)
then wat is f^(-1)(x)?

i first took the ln of both sides....

getting lny = ln (2e^x - 8)/(10e^x + 9)

then using one of the properties i get

lny = ln (2e^x - 8) - ln (10e^x + 9)

and from here i get stuck...how do i solve for x?? am i doing it a totally wrong way?? plz help!!!!!

and i did get a sol'n from sum1 but i had a question abt it.......some1 b kind enuf 2 help!!!

[Astrophysics]

f^-1 = 1/f

1/((2e^x -8)/(10e^x + 9)) = (10e^x + 9)/(2e^x -8)

then you can work to write this function simpler.

[KLscilevothma]

f-1 is the inverse of f, where f is a function.

f-1 does not equal 1/f

[Astrophysics]

if x^-1 = 1/x
it would mean the same for a function, so

f^-1 would be the same as 1/f

[MathNerd]

Originally posted by Astrophysics
if x^-1 = 1/x
it would mean the same for a function, so

f^-1 would be the same as 1/f Sorry, you're wrong Astrophysics. If f is a function then f^-1 is the inverse of the function so that f^-1(f(x)) = x.

[MathNerd]

Originally posted by sonya
ok....since no1s helpin in the hmwk section thot id take a shot here..

if f(x)= (2e^x -8)/(10e^x + 9)
then wat is f^(-1)(x)?

i first took the ln of both sides....

getting lny = ln (2e^x - 8)/(10e^x + 9)

then using one of the properties i get

lny = ln (2e^x - 8) - ln (10e^x + 9)

and from here i get stuck...how do i solve for x?? am i doing it a totally wrong way?? plz help!!!!!

and i did get a sol'n from sum1 but i had a question abt it.......some1 b kind enuf 2 help!!!

You just need some algebraic manipulation first

f = (2e^x -8)/(10e^x + 9)
so, 10e^x * f + 9f = 2e^x -8
so, 10e^x * f - 2e^x = -8 - 9f
so, e^x ( 10f - 2 ) = -( 8 + 9f )
so, e^x = -( 8 + 9f ) / ( 10f - 2 )
so, x = ln[ ( 8 + 9f ) / ( 2 - 10f ) ]
so, x = ln( 8 + 9f ) - ln( 2 - 10f )

therefore f^-1(x) = ln( 8 + 9x ) - ln( 2 - 10x )

[Astrophysics]

Originally posted by MathNerd
Sorry, you're wrong Astrophysics. If f is a function then f^-1 is the inverse of the function so that f^-1(f(x)) = x.


oh thanks mathnerd, but...how come when I put this in my calculator I get f^-1 = 1/f and f^-1(f(x)) is not x??????

[MathNerd]

Originally posted by Astrophysics
oh thanks mathnerd, but...how come when I put this in my calculator I get f^-1 = 1/f and f^-1(f(x)) is not x??????

Because your calculator thinks you're talking about the multiplicative inverse. It doesn't realize that you really want the inverse of the function. In mathematical notation f^-1(f(x)) = f(f^-1(x)) = x e.g. cos^-1 x does not equal 1 / cos x, it is instead inverse cos which is the inverse function of the cosine function. In my last post where I solved for f^-1(x), you can easily show that f(f^-1(x)) = f^-1(f(x)) = x

[Astrophysics]

I'm very confused here, I was actually thinking the same thing as my calculator I guess, but I have no idea what these inverse means, is it f^-1 or is it something like f' ???

[MathematicalPhysicist]

Originally posted by MathNerd
Because your calculator thinks you're talking about the multiplicative inverse. It doesn't realize that you really want the inverse of the function. In mathematical notation f^-1(f(x)) = f(f^-1(x)) = x e.g. cos^-1 x does not equal 1 / cos x, it is instead inverse cos which is the inverse function of the cosine function. In my last post where I solved for f^-1(x), you can easily show that f(f^-1(x)) = f^-1(f(x)) = x
or you could say cos^-1 x=arccos x

[MathNerd]

Originally posted by loop quantum gravity
or you could say cos^-1 x=arccos x Yes, well ArcCos is nothing but the inverse of the cosine function, whereas 1/cos(x) = sec(x), it is easily shown that sec(x) doesn't equal ArcCos(x). By definition the inverse of an arbitary function f(x) is another function g(x) such that g(f(x)) = f(g(x)) = x. In mathematical notation g(x) = f^-1(x)

[phoenixthoth]

inverses and identities.

unification.

multiplication.
1 is the "multiplicative identity." let x be a nonzero real number.

x * x^-1 = x^-1 * x = 1. in effect, the thing and its inverse cancel, leaving the identity.

functions.
let * stand for the composition of functions. so rather than write
f(g(x)), write f * g (x).

f * f^-1 = f^-1 * f = I. by, "I", i mean the "identity function,"
I(x) = x.

except for zero, all numbers have a mulitplicative inverse. not so for functions. for example, the function f(x) = x^2 has no inverse defined on the set of real numbers. however, if restriced to nonnegative numbers, f does have an inverse, namely the square root of x. that is precisely what will "cancel" or "oppose" the operation of squaring.

cheers,
phoenix