Question

f(x) = 19xe^(1 − x^2)
Find the exact location of all the relative and absolute extrema of the function

@ganeshie8

Answer 1

same thing
1) find the first derivative
2) set it equal to 0 and solve x

Answer 2

19e^(1-x^2) (-2x)?

Answer 3

you need to use product rule and chainrule

Answer 4

wolfram gives this
http://www.wolframalpha.com/input/?i=%2819xe%5E%281+%E2%88%92+x%5E2%29%29%27

Answer 5

got it

Answer 6

so now set it = to 0??

Answer 7

yes

Answer 8

\[\large f'(x) = -19e^{1-x^2}(2x^2-1) = 0\]

Answer 9

obviously 19 is not 0

Answer 10

can `e^something` ever become 0 ?

Answer 11

im lost

Answer 12

is it 1/squareroot 2

Answer 13

you should get +- 1/sqrt(2)

Answer 14

two points

Answer 15

so my answer is just 1/sqrt2,-1/sqrt2?

Answer 16

nope, those are the x values

Answer 17

plugin each value into f(x) for y coordinate

Answer 18

for the positive i got 23.4942781075

Answer 19

doesnt look correct

Answer 20

>.<

Answer 21

yea its wrong

Answer 22

use wolfram for calculations
http://www.wolframalpha.com/input/?i=evaluate+19xe%5E%281+%E2%88%92+x%5E2%29+at+x%3D1%2Fsqrt%282%29

Answer 23

so it is saying that it is wrong

Answer 24

wolfram is giving y = 22.1506 when x = 1/sqrt(2) right ?

Answer 25

yes

Answer 26

so one critical point is (1/sqrt(2) , 22.1506)

Answer 27

maximum

Answer 28

yes

Answer 29

but it is saying its wrong

Answer 30

the minimum wold be (-1/sqrt(2) , - 22.1506)

Answer 31

it still wrong>.<

Answer 32

try entering `(0.71, 22.15)` for maximum

Answer 33

online graders are usually dumb
you need to spoonfeed them in the exact format they digest

Answer 34

nope it didnt take it

Answer 35

take a screenshot and attach

Answer 36

\[x = \pm \frac{ 1 }{ \sqrt{2} }\]

does that work?

Answer 37

What ganeshie said is correct, maybe rounding errors or something?