ellieb34
Find all values of x for which the function y=(2(8-5x))/(x+1)
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mukushla
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i think question is incomplete !
ellieb34
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is differentiable.
ellieb34
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y=([\sqrt{8-5x}\])/(x+1)...it's a square root also..not a two
mukushla
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\[f(x)=\frac{\sqrt{8-5x}}{x+1}\]
ellieb34
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yes, that's what it looks like!
mukushla
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ok u must check for continuity first
ellieb34
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when I graphed it, it looked like 1/x graph
mukushla
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yeah its something like that..maybe
ellieb34
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so how do I check for continuity?
mukushla
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ok u have some restrictions here..
division by zero is not allowed
under radical cant be negative
mukushla
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actually u must find domain of ur function
ellieb34
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domain is x=-1
mukushla
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\[x\neq -1\]
ellieb34
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ohyeah you're right
mukushla
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and what about radical?
ellieb34
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what's radical?
mukushla
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square root
ellieb34
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x\[\neq5/8\]
ellieb34
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er 8/5
mukushla
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when u have something like \[\sqrt{x}\] u must have \[x\ge0\] right ?
ellieb34
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yes
mukushla
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so what about the expression under square root sign?
ellieb34
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\[x \neq 8/5\]
mukushla
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emm i mean u must have \(8-5x\ge0\) and this leads to\[x\le\frac{8}{5}\]make sense?
ellieb34
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oh yes sorry i meant to do that sign instead of the nonequal sign
ellieb34
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it does make sense
mukushla
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good
mukushla
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ok so the domain of function becomes\[x\le \frac{8}{5} \ \ \text{and} \ \ x\neq-1\]
mukushla
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or u can write it like this\[(-\infty,-1)\cup (-1,\frac{8}{5}]\]
ellieb34
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got it. so is that where it's differentiable?
mukushla
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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...
ellieb34
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ok thank you so much! i appreciate it :)
mukushla
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very welcome :)