## mathcalculus 2 years ago Help (Attached below) Determine the extrema of below on the given interval

1. mathcalculus

2. mathcalculus

i know we have to find the derivative and set it to zero

3. mathcalculus

i got x= 8 and x= 2/9

4. mathcalculus

now to have the intervals, im a little lost on...

5. mathcalculus

would there only be 2 intervals?

6. mathcalculus

or 3? 0-2/9 then 2/9-8 and 8-infinity?

7. campbell_st

ok... so whats the derivative...?

8. mathcalculus

9x^2-74x+16

9. mathcalculus

then i set it to = 0

10. campbell_st

ok... so what are the solutions...?

11. mathcalculus

x=8 and x=2/9

12. mathcalculus

critical numbers

13. campbell_st

yep... so on [0, 4] substitute x = 0, 2/9 and 4 into the original equation and find the corresponding values...

14. mathcalculus

why 4?

15. mathcalculus

thanks

16. mathcalculus

isn't 4 a y?

17. mathcalculus

and what 8?

18. mathcalculus

because the 2 critical numbers are 8 and 2/9

19. campbell_st

well (a) is asking for the location of that min and max on the interval x = 0 to x = 4 and as x = 2/9 is a critical point...you don't know its nature

20. mathcalculus

ooooooh, 8 would be the other one

21. mathcalculus

let me try it

22. campbell_st

so the value x =0 is f(0) = 2 x = 2/9 f(2/9) = 3.76 x = 4 f(4) = ?

23. mathcalculus

-334

24. mathcalculus

so the max is 3.76 and min is -334?

25. campbell_st

thats correct...

26. campbell_st

so just test x = -9, x = 8 and x = 9... and see what happens... I think x = 2/9 remains the max

27. campbell_st

its an add interval since x = -9 gives a very large negative number