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i think question is incomplete !

is differentiable.

y=([\sqrt{8-5x}\])/(x+1)...it's a square root also..not a two

\[f(x)=\frac{\sqrt{8-5x}}{x+1}\]

yes, that's what it looks like!

ok u must check for continuity first

when I graphed it, it looked like 1/x graph

yeah its something like that..maybe

so how do I check for continuity?

ok u have some restrictions here..
division by zero is not allowed
under radical cant be negative

actually u must find domain of ur function

domain is x=-1

\[x\neq -1\]

ohyeah you're right

and what about radical?

what's radical?

square root

x\[\neq5/8\]

er 8/5

when u have something like \[\sqrt{x}\] u must have \[x\ge0\] right ?

yes

so what about the expression under square root sign?

\[x \neq 8/5\]

emm i mean u must have \(8-5x\ge0\) and this leads to\[x\le\frac{8}{5}\]make sense?

oh yes sorry i meant to do that sign instead of the nonequal sign

it does make sense

good

ok so the domain of function becomes\[x\le \frac{8}{5} \ \ \text{and} \ \ x\neq-1\]

or u can write it like this\[(-\infty,-1)\cup (-1,\frac{8}{5}]\]

got it. so is that where it's differentiable?

ok thank you so much! i appreciate it :)

very welcome :)