anonymous
  • anonymous
Help (Attached below) Determine the extrema of below on the given interval
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
anonymous
  • anonymous
i know we have to find the derivative and set it to zero
anonymous
  • anonymous
i got x= 8 and x= 2/9

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
now to have the intervals, im a little lost on...
anonymous
  • anonymous
would there only be 2 intervals?
anonymous
  • anonymous
or 3? 0-2/9 then 2/9-8 and 8-infinity?
campbell_st
  • campbell_st
ok... so whats the derivative...?
anonymous
  • anonymous
9x^2-74x+16
anonymous
  • anonymous
then i set it to = 0
campbell_st
  • campbell_st
ok... so what are the solutions...?
anonymous
  • anonymous
x=8 and x=2/9
anonymous
  • anonymous
critical numbers
campbell_st
  • campbell_st
yep... so on [0, 4] substitute x = 0, 2/9 and 4 into the original equation and find the corresponding values...
anonymous
  • anonymous
why 4?
anonymous
  • anonymous
thanks
anonymous
  • anonymous
isn't 4 a y?
anonymous
  • anonymous
and what 8?
anonymous
  • anonymous
because the 2 critical numbers are 8 and 2/9
campbell_st
  • 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
anonymous
  • anonymous
ooooooh, 8 would be the other one
anonymous
  • anonymous
let me try it
campbell_st
  • campbell_st
so the value x =0 is f(0) = 2 x = 2/9 f(2/9) = 3.76 x = 4 f(4) = ?
anonymous
  • anonymous
-334
anonymous
  • anonymous
so the max is 3.76 and min is -334?
campbell_st
  • campbell_st
thats correct...
campbell_st
  • campbell_st
so just test x = -9, x = 8 and x = 9... and see what happens... I think x = 2/9 remains the max
campbell_st
  • campbell_st
its an add interval since x = -9 gives a very large negative number

Looking for something else?

Not the answer you are looking for? Search for more explanations.