Question

Can someone help me, i am trying to find the abso. and rel. max and min of :y=-x^2+6x+7, over the interval [0,5]

Answer 1

I know that i first have to take the derivative, and then factor it. But i get x=-3. am i procedding the right way?

Answer 2

Yes diff, then set dy/dx=0 for all of the turning pts.

Answer 3

So I get x=+3

Answer 4

right, now i also have to check the endpoints of the intervals correct?

Answer 5

Yes

Answer 6

thanks

Answer 7

one more question please, when i check the endpoint do i plug the 0 and the 5, into the the orginal function?

Answer 8

yes, the original function. your trying to check the min/max of the original function, the derivative is just a tool that helps you accomplish that.

Answer 9

Also, when i take the derivative of the function i posted, i get, y'=-2(x-3). I kgot x=3, and i plugged into the original function and got 16. This would be considered a a max, right, since it when from + to 0 to -? Now would this also be considered a abso. max?

Answer 10

would it be considered a absolute max i mean?

Answer 11

yes, you are correct :)

Answer 12

so, when i am dealing with an interval and i get a relative max, at an x value that lies bewteen the interval it is also considered a aboslute max?

Answer 13

Could you help me find the rel/abso min?

Answer 14

This is where the definitions get a bit fuzzy >.< it is an absolute max for that interval., it might not be the absolute max for the function, but you are only concerned with [0,5]

Answer 15

for the minimum, you should look at the end points x = 0 and x = 5. Because your derivative only returned one solution (which was a max), the min cant be inside the interval.

Answer 16

but, if i get a problem that asks me to evalutae the rel/abso max min for an interval [a,b], then once i find a rel max that is within [a,b] it would also be considered a abso, max correct?

Answer 17

yes. you are correct. just worry about your interval, ignore everything else.

Answer 18

and vice versa right, if i get am dealing with [a,b] screnario of a function, and i get a max min at an endpoint it is also a rel max and min or no?

Answer 19

When i plug in 0, into my original function, i get 7, now how do i know wheter this is a max or min? thanks you

Answer 20

hmm...here is a scenario where you have relative and absolute max's. Lets say you have a function, you take the first derivative, and it gives you 2 points: x = 3, x = 5, and both are in your interval. Then you plug those into the function, and get f(3) = 20, and f(5) = 25. Plugging your end points (interval points) into the function you get f(a) = 10, f(b) = -20 (interval is [a,b]).

Answer 21

25 is absolute max. 20 is relative max (local max is what i call it lol), -20 is absolute min, 10 is relative min

Answer 22

could you help me with the function i posted, its just that i want o solve this one, and then i think from there i can handle the rest

Answer 23

I found the relitave max, i belive , 16, but i need to evaluate the function at the endpoint still

Answer 24

so you plugged in 0, and got 7, that has to be a min

Answer 25

when you plug in 5 you get...12, so now we know that 7 is the absolute min, and 12 is a relative min

Answer 26

how can i make sure for instance that 0 is a abs. min?

Answer 27

I am not saying that you are wrong, i just want to be able to prove tha 0 is indded and abso. min

Answer 28

if x = 0 didnt produce a min, then inbetween x = 0 and x = 3 there would be a part of the function lower than it. But if there was a part lower than it, it would have to come back up to reach it, creating a "valley" in the function. When you take the derivative and set it equal to 0, it tells you where the "hills" and "valleys" are, and there was only one hill (at x =3)

Answer 29

if there were points lower than f(0) = 7, your derivative would have told you.

Answer 30

that makes alot of sence thank youi appreciate you time and patience

Answer 31

no prob :)