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Mathematics 9 Online
OpenStudy (lxelle):

A function f is such that f(x) = 15/ (2x+3) for 0 <=x<=6. Find an expression for f'(x) and use your result to explain why f has an inverse.

OpenStudy (gorv):

\[\frac{ d f(x)}{ dx}\]

OpenStudy (lxelle):

f'(x) = -30 (2x+3)^-2 How do you know whether f has an inverse or not. It's something go to do with the turning point.

OpenStudy (gorv):

u know how to differentiate ??

OpenStudy (gorv):

well the power which was before differentiation will come down

OpenStudy (gorv):

so -1 will come down and multiplied with 15

OpenStudy (lxelle):

What? Lol.

OpenStudy (lxelle):

I already differentiated.

OpenStudy (gorv):

\[\frac{ 15 }{ x+3 }=15*(x+3)^{-1}\]

OpenStudy (lxelle):

Please check before you post.

OpenStudy (gorv):

-15*(x+3)^(-2)

OpenStudy (gorv):

yeah dear i already checked

OpenStudy (gorv):

got me now ??

OpenStudy (lxelle):

Then? Explain why f has an inverse. I know how to differentiate. I just need to know why f has an inverse!!!!

OpenStudy (gorv):

a function should be one one ..u know abt it ??

OpenStudy (lxelle):

Yes, but what does it got to do with the turning points?

OpenStudy (gorv):

hmm...we need to check for turning point because at that point inverse may exist or may not

OpenStudy (lxelle):

It's so frustrating talking to you. You're not answering my question!

OpenStudy (gorv):

loll sorry..i was solving other Q and didnt recieve notificatrion...sorrrry dear

OpenStudy (gorv):

yeah tell me now i m here

OpenStudy (gorv):

generally we dont check for turning point

OpenStudy (gorv):

if function is one one its ionverse will exist

OpenStudy (lxelle):

No. I mean, why is it that no turning point = an inverse?

OpenStudy (gorv):

becozz at turning point inverse wont exist

OpenStudy (gorv):

turning point will not be in our domain

OpenStudy (gorv):

15/(2x+3) here we can not take x=-3/2 because it will make denominator zero

OpenStudy (gorv):

and denominator should not be zero for a function

OpenStudy (gorv):

bcozz it will make value of function infinity

OpenStudy (gorv):

any more doubt ???

OpenStudy (lxelle):

yes. i have doubt, as i dont understand what you're trying to say. haha. can you give me examples?

OpenStudy (gorv):

hmm ...like f(x)=1/x

OpenStudy (gorv):

what will be its domain ??

OpenStudy (gorv):

like here we cant put x=0

OpenStudy (gorv):

bcozz 1/0= infinity

OpenStudy (gorv):

but value of function should be finite

OpenStudy (gorv):

so x=0 is turning point

OpenStudy (lxelle):

omg, can give simpler one. Sorry. I still can't catch.

OpenStudy (lxelle):

@amistre64

OpenStudy (amistre64):

turning point is a fair way to assess an inverse. for example: y=x^2 has a turning point, and has no inverse since: (2)^2 = 4 and (-2)^2 = 4 an inverse would mean that if we wanted to know what makes y=4, there is more than one solution

OpenStudy (amistre64):

alot to read over here, what is your question?

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