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.
\[\frac{ d f(x)}{ dx}\]
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.
u know how to differentiate ??
well the power which was before differentiation will come down
so -1 will come down and multiplied with 15
What? Lol.
I already differentiated.
\[\frac{ 15 }{ x+3 }=15*(x+3)^{-1}\]
Please check before you post.
-15*(x+3)^(-2)
yeah dear i already checked
got me now ??
Then? Explain why f has an inverse. I know how to differentiate. I just need to know why f has an inverse!!!!
a function should be one one ..u know abt it ??
Yes, but what does it got to do with the turning points?
hmm...we need to check for turning point because at that point inverse may exist or may not
It's so frustrating talking to you. You're not answering my question!
loll sorry..i was solving other Q and didnt recieve notificatrion...sorrrry dear
yeah tell me now i m here
generally we dont check for turning point
if function is one one its ionverse will exist
No. I mean, why is it that no turning point = an inverse?
becozz at turning point inverse wont exist
turning point will not be in our domain
15/(2x+3) here we can not take x=-3/2 because it will make denominator zero
and denominator should not be zero for a function
bcozz it will make value of function infinity
any more doubt ???
yes. i have doubt, as i dont understand what you're trying to say. haha. can you give me examples?
hmm ...like f(x)=1/x
what will be its domain ??
like here we cant put x=0
bcozz 1/0= infinity
but value of function should be finite
so x=0 is turning point
omg, can give simpler one. Sorry. I still can't catch.
@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
alot to read over here, what is your question?
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