Inverse function URGENT!!!!!

Hi can anyone help me with this inverse question please, this is what I've done so far

ImageUploadedByStudent Room1413550368.769142.jpg

I just don't know where to go from here.

Much appreciated :smile:

Posted from TSR Mobile

Reply 1

username1050473

Could you try multiplying the fraction by

1 + e^x

? (completing the square)

Reply 2

Smaug123

Original post by puddinboy

Hi can anyone help me with this inverse question please, this is what I've done so far

ImageUploadedByStudent Room1413550368.769142.jpg

I just don't know where to go from here.

Much appreciated :smile:

Posted from TSR Mobile

As sarcasmrules implies, substitute u=e^x and solve for u.

Reply 3

username1050473

Original post by Smaug123

As sarcasmrules implies, substitute u=e^x and solve for u.

Why would you need to substitute

u = e^x

Reply 4

TeeEm

Original post by puddinboy

Hi can anyone help me with this inverse question please, this is what I've done so far

ImageUploadedByStudent Room1413550368.769142.jpg

I just don't know where to go from here.

Much appreciated :smile:

Posted from TSR Mobile

This looked quite an easy question to start with until I tried it ...

I managed to get a rearrangement but this yields plus or minus which
implies the function is not one to one. It is defined for x>0 but it is neither increasing or decreasing.

Do you have the printed version?

Reply 5

Hasufel

do as both the posters above suggested, so when you group everything on the side where

e^{2x}

is +ve,

you will have a disguised quadratic in

e^{x}

i.e something of the form

(e^{x})^{2}+p(e^{x})+q=0

, p and q both functions of y

solve this for e^x with the quadratic formula, noting that you can only have ONE solution, since for the discriminant, y has to be greater than or equal to....

(try differentiating - if you dare! and take the numerator as=0, then plug into H(x) and see what you get)

then take logs

(EDIT)

(edited 9 years ago)

Reply 6

TeeEm

Original post by Hasufel

do as both the posters above suggested, so when you group everything on the side where

e^{2x}

is +ve,

you will have a disguised quadratic in

e^{x}

i.e something of the form

(e^{x})^{2}+p(e^{x})+q=0

solve this for e^x with the quadratic formula, noting that you can only have ONE solution, since for the discriminant, y has to be greater than....

then take logs

he is probably attempting this now but what about the domain?

this is not a one to one function.

Reply 7

Hasufel

Original post by TeeEm

he is probably attempting this now but what about the domain?

this is not a one to one function.

Hmm...you're right!....

Reply 8

Hasufel

Original post by Hasufel

Hmm...you're right!....

domain would be, for the inverse, the minimum point for the H(x), as corresponding to the x value below this, the inverse function is undefined (when you solve you get the horrible root thing, then for domain , set = 1 and solve ***since ln(1) is undefined) - this has the same value as the minumum point for H(x)

*** for this bit, i cheated and used Maple, but for the differentiation of H(x) itself, i did it by hand and got the same answer.

Naturally, then if the range of H is greater than or equal to this, the domain of the inverse has to be as well.

(edited 9 years ago)

Reply 9

TeeEm

Original post by Hasufel

domain would be, for the inverse, the minimum point for the H(x), as below this, the inverse function is undefined (when you solve you get the horrible root thing, then for domain , set = 1 and solve ***since ln(1) is undefined) - this has the same value as the minumum point for H(x)

*** for this bit, i cheated and used Maple, but for the differentiation, i did it by hand and got the same answer.

Original post by puddinboy

Hi can anyone help me with this inverse question please, this is what I've done so far

ImageUploadedByStudent Room1413550368.769142.jpg

I just don't know where to go from here.

Much appreciated :smile:

Posted from TSR Mobile

this is a good question and I am going to steal it and adapt it ...
The hints and ideas of the previous "TSR posters" are sensible.

I will supply an answer (hopefully correct) for you to check.

ANSWER TO FUNCTION INVERSE, NO SOLUTION.pdf

Reply 10

Hasufel