Sum1 here help

  • Thread starter sonya
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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!!!
 
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.
 
f-1 is the inverse of f, where f is a function.

f-1 does not equal 1/f
 
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.
 
Last edited by a moderator:

MathNerd

Re: sum1 here help!!!

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 )
 
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
 
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

Gold Member
4,118
145
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)
 
Last edited by a moderator:
1,570
1
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
 

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