Question

Find the best possible bounds fo rthe following equation:
y=e^-(x)^2 for absolute x which is less than or equal to .3

Answer 1

it is .3

Answer 2

\[y=e^{-x^2} ,\left| x \right| \le0.3\]

Answer 3

ya

Answer 4

I'm not sure what you mean by best possible bounds

Answer 5

do u know abt lower bounds and upper bounds?

Answer 6

cus i don;t!!!!

Answer 7

I think it is in other words the yvalue of the global min and max

Answer 8

upperbound is global max and lower bound is global min

Answer 9

Right that is pretty much my understanding.

Answer 10

i guess you would find the critical points and the end points which is .3

Answer 11

Right, so we differentiate the above to get\[{dy \over dx}=-2xe^{-x^2}=0\to x=\left\{ 0,\infty \right\}\]only three points to check then, our critical point and the endpoints:
f(-0.3) f(0) and f(0.3)

f(-0.3)=e^0.9=2.46
f(0)=1
f(0.3)=e^-0.9=0.41

so I guess the bounds are those numbers...

Answer 12

2.46 and 0.41
I mean

Answer 13

sorry i was gone i just came back

Answer 14

I dunno if this is what you want, but it's all I can think to do. I still don't know what it means by "best bounds" though.

Answer 15

i mean the global min and global max

Answer 16

well these are the global max and min within those bounds, so there you have it

Answer 17

there shld not be a -.3

Answer 18

since it is absolute x

Answer 19

I have a question