Sum1 here help

1. Sep 13, 2003

sonya

sum1 here help!!!

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!!!

2. Sep 13, 2003

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.

3. Sep 13, 2003

KLscilevothma

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

f-1 does not equal 1/f

4. Sep 13, 2003

Astrophysics

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

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

5. Sep 13, 2003

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.

Last edited by a moderator: Sep 13, 2003
6. Sep 13, 2003

MathNerd

Re: sum1 here 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 )

7. Sep 13, 2003

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??????

8. Sep 13, 2003

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

9. Sep 13, 2003

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' ???

10. Sep 13, 2003

MathematicalPhysicist

or you could say cos^-1 x=arccos x

11. Sep 13, 2003

MathNerd

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)

Last edited by a moderator: Sep 13, 2003
12. Sep 13, 2003

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