anonymous
  • anonymous
Find all values of x for which the function y=(2(8-5x))/(x+1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i think question is incomplete !
anonymous
  • anonymous
is differentiable.
anonymous
  • anonymous
y=([\sqrt{8-5x}\])/(x+1)...it's a square root also..not a two

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anonymous
  • anonymous
\[f(x)=\frac{\sqrt{8-5x}}{x+1}\]
anonymous
  • anonymous
yes, that's what it looks like!
anonymous
  • anonymous
ok u must check for continuity first
anonymous
  • anonymous
when I graphed it, it looked like 1/x graph
anonymous
  • anonymous
yeah its something like that..maybe
anonymous
  • anonymous
so how do I check for continuity?
anonymous
  • anonymous
ok u have some restrictions here.. division by zero is not allowed under radical cant be negative
anonymous
  • anonymous
actually u must find domain of ur function
anonymous
  • anonymous
domain is x=-1
anonymous
  • anonymous
\[x\neq -1\]
anonymous
  • anonymous
ohyeah you're right
anonymous
  • anonymous
and what about radical?
anonymous
  • anonymous
what's radical?
anonymous
  • anonymous
square root
anonymous
  • anonymous
x\[\neq5/8\]
anonymous
  • anonymous
er 8/5
anonymous
  • anonymous
when u have something like \[\sqrt{x}\] u must have \[x\ge0\] right ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so what about the expression under square root sign?
anonymous
  • anonymous
\[x \neq 8/5\]
anonymous
  • anonymous
emm i mean u must have \(8-5x\ge0\) and this leads to\[x\le\frac{8}{5}\]make sense?
anonymous
  • anonymous
oh yes sorry i meant to do that sign instead of the nonequal sign
anonymous
  • anonymous
it does make sense
anonymous
  • anonymous
good
anonymous
  • anonymous
ok so the domain of function becomes\[x\le \frac{8}{5} \ \ \text{and} \ \ x\neq-1\]
anonymous
  • anonymous
or u can write it like this\[(-\infty,-1)\cup (-1,\frac{8}{5}]\]
anonymous
  • anonymous
got it. so is that where it's differentiable?
anonymous
  • anonymous
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...
anonymous
  • anonymous
ok thank you so much! i appreciate it :)
anonymous
  • anonymous
very welcome :)

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