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ellieb34
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Find all values of x for which the function y=(2(85x))/(x+1)
 one year ago
 one year ago
ellieb34 Group Title
Find all values of x for which the function y=(2(85x))/(x+1)
 one year ago
 one year ago

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mukushla Group TitleBest ResponseYou've already chosen the best response.0
i think question is incomplete !
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
is differentiable.
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
y=([\sqrt{85x}\])/(x+1)...it's a square root also..not a two
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
\[f(x)=\frac{\sqrt{85x}}{x+1}\]
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
yes, that's what it looks like!
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
ok u must check for continuity first
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
when I graphed it, it looked like 1/x graph
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
yeah its something like that..maybe
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
so how do I check for continuity?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
ok u have some restrictions here.. division by zero is not allowed under radical cant be negative
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
actually u must find domain of ur function
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
domain is x=1
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
\[x\neq 1\]
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
ohyeah you're right
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
and what about radical?
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
what's radical?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
square root
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
x\[\neq5/8\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
when u have something like \[\sqrt{x}\] u must have \[x\ge0\] right ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
so what about the expression under square root sign?
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
\[x \neq 8/5\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
emm i mean u must have \(85x\ge0\) and this leads to\[x\le\frac{8}{5}\]make sense?
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
oh yes sorry i meant to do that sign instead of the nonequal sign
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
it does make sense
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
ok so the domain of function becomes\[x\le \frac{8}{5} \ \ \text{and} \ \ x\neq1\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
or u can write it like this\[(\infty,1)\cup (1,\frac{8}{5}]\]
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
got it. so is that where it's differentiable?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
yes thats it ... and just one more point function is not continous at end points \(x=1\) and \(x=\frac{8}{5}\) so it'll not be differentiable at that points...
 one year ago

ellieb34 Group TitleBest ResponseYou've already chosen the best response.0
ok thank you so much! i appreciate it :)
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
very welcome :)
 one year ago
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