## ellieb34 2 years ago Find all values of x for which the function y=(2(8-5x))/(x+1)

1. mukushla

i think question is incomplete !

2. ellieb34

is differentiable.

3. ellieb34

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

4. mukushla

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

5. ellieb34

yes, that's what it looks like!

6. mukushla

ok u must check for continuity first

7. ellieb34

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

8. mukushla

yeah its something like that..maybe

9. ellieb34

so how do I check for continuity?

10. mukushla

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

11. mukushla

actually u must find domain of ur function

12. ellieb34

domain is x=-1

13. mukushla

$x\neq -1$

14. ellieb34

ohyeah you're right

15. mukushla

16. ellieb34

17. mukushla

square root

18. ellieb34

x$\neq5/8$

19. ellieb34

er 8/5

20. mukushla

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

21. ellieb34

yes

22. mukushla

so what about the expression under square root sign?

23. ellieb34

$x \neq 8/5$

24. mukushla

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

25. ellieb34

26. ellieb34

it does make sense

27. mukushla

good

28. mukushla

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

29. mukushla

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

30. ellieb34

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

31. mukushla

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...

32. ellieb34

ok thank you so much! i appreciate it :)

33. mukushla

very welcome :)