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Stationary points core 3 Watch

1. stauionary points on the curve y = x^2 e ^ -x
y' = xe^-x

so x=0 and x=2

y"= 2x(e^-x) + - e^-x(2)

at 0,0 max point

at 2,0.541 min point

i have done the working out for finding min and max just cba typing it out, just simplified everything down for here
2. (Original post by Custardcream000)
stauionary points on the curve y = x^2 e ^ -x
y' = xe^-x

so x=0 and x=2

y"= 2x(e^-x) + - e^-x(2)

at 0,0 max point

at 2,0.541 min point

i have done the working out for finding min and max just cba typing it out, just simplified everything down for here
3. (Original post by Custardcream000)
stauionary points on the curve y = x^2 e ^ -x
y' = xe^-x

so x=0 and x=2

y"= 2x(e^-x) + - e^-x(2)

at 0,0 max point

at 2,0.541 min point

i have done the working out for finding min and max just cba typing it out, just simplified everything down for here
Don't use decimals. Leave it in exact form.
4. (Original post by Zacken)
find the stationary points on the curve...
5. (Original post by RDKGames)
Don't use decimals. Leave it in exact form.
x=2, y = 2^2(e^-2) = 0.5414311329 there is no exact form
6. (Original post by Custardcream000)
find the stationary points on the curve...
Your first derivative is wrong.
7. (Original post by Custardcream000)
x=2, y = 2^2(e^-2) = 0.5414311329 there is no exact form
is the exact form.... you've literally written it out...

...
8. (Original post by Custardcream000)
x=2, y = 2^2(e^-2) = 0.5414311329 there is no exact form
..........

Yes there is. You literally wrote it out right after the equals sign on y.

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