Qualification > Math
Additional Math Help HERE ONLY...!
cooldude:
--- Quote from: Ari Ben Canaan on May 30, 2010, 10:26:33 am ---Sorry Darren, but this is way out of my depth.
I do normal mathematics. Wish I could help you out.
Just hang in there; someone will be along to help you out ;)
--- End quote ---
l) okay substitute the value of 0 in the function-->
therefore f(0)=e^-1=0.368
therefore f(x)>0.368
we substitute the value of 0 because this is the least value in the domain, i know that 0 is not included, you can take something like 0.000000000000000001, but 0 is better. or we can use differentiation to find the minimum value of the function, d/dx (e^(x-1))=e^(x-1), therefore at 0 it is 0.368, therefore since this is the minimum value every value of y will be greater than 0.368, therefore the range is f(x)>0.368
ii) let f(x)=y,
now y=e^(x-1)
take log on both sides (to find the inverse we have to make x the subject)
ln y=x-1
x=ln y + 1
therefore f^-1(x)=ln x + 1
iii) the domain of f^-1(x)=range of f(x), therefore the domain is x>0.368
J.Darren:
--- Quote from: cooldude on May 30, 2010, 10:55:26 am ---l) okay substitute the value of 0 in the function-->
therefore f(0)=e^-1=0.368
therefore f(x)>0.368
we substitute the value of 0 because this is the least value in the domain, i know that 0 is not included, you can take something like 0.000000000000000001, but 0 is better. or we can use differentiation to find the minimum value of the function, d/dx (e^(x-1))=e^(x-1), therefore at 0 it is 0.368, therefore since this is the minimum value every value of y will be greater than 0.368, therefore the range is f(x)>0.368
ii) let f(x)=y,
now y=e^(x-1)
take log on both sides (to find the inverse we have to make x the subject)
ln y=x-1
x=ln y + 1
therefore f^-1(x)=ln x + 1
iii) the domain of f^-1(x)=range of f(x), therefore the domain is x>0.368
--- End quote ---
Thanks mate, although I don't quite get the final bit RE: domain of the inverse function = range of the function ...
By the way what would the inverse function look like if the original function is, say for example, ln (x-1) ?
J.Darren:
By the way guys I have posted a list containing the rules of differntiation and integration, would you be kind enough to proof-read it to ensure that I have not left out anything ?
https://studentforums.biz/index.php/topic,8457.0.html
Ghost Of Highbury:
--- Quote from: J.Darren on May 30, 2010, 11:08:09 am ---Thanks mate, although I don't quite get the final bit RE: domain of the inverse function = range of the function ...
By the way what would the inverse function look like if the original function is, say for example, ln (x-1) ?
--- End quote ---
the inverse of any graph f(x) is the graph under reflection along y=x.
cooldude:
--- Quote from: J.Darren on May 30, 2010, 11:08:09 am ---Thanks mate, although I don't quite get the final bit RE: domain of the inverse function = range of the function ...
By the way what would the inverse function look like if the original function is, say for example, ln (x-1) ?
--- End quote ---
yep, the same applies (vice-versa) --> the range of an inverse function = to the domain of the function
and the reason is given by aadi
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