## anonymous 3 years ago Find all values of x for which the function y=(2(8-5x))/(x+1)

1. anonymous

i think question is incomplete !

2. anonymous

is differentiable.

3. anonymous

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

4. anonymous

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

5. anonymous

yes, that's what it looks like!

6. anonymous

ok u must check for continuity first

7. anonymous

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

8. anonymous

yeah its something like that..maybe

9. anonymous

so how do I check for continuity?

10. anonymous

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

11. anonymous

actually u must find domain of ur function

12. anonymous

domain is x=-1

13. anonymous

$x\neq -1$

14. anonymous

ohyeah you're right

15. anonymous

16. anonymous

17. anonymous

square root

18. anonymous

x$\neq5/8$

19. anonymous

er 8/5

20. anonymous

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

21. anonymous

yes

22. anonymous

so what about the expression under square root sign?

23. anonymous

$x \neq 8/5$

24. anonymous

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

25. anonymous

26. anonymous

it does make sense

27. anonymous

good

28. anonymous

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

29. anonymous

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

30. anonymous

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

31. 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...

32. anonymous

ok thank you so much! i appreciate it :)

33. anonymous

very welcome :)