Question

Find the extrema and the points of inflection (if any exist) of the function. Use a graphing utility to graph the function and confirm your results. (Round your answers to three decimal places. Order your answers from smallest to largest x, then from smallest to largest y.)

g(x) = 1/√2π^(e−(x − 5)2/2)

Answer 1

IT looks like that.

Answer 2

2pi is under the sqrt.

Answer 3

it's a weird looking problem lol.

Answer 4

\[\huge \frac1{\sqrt{2\pi}^{\left(e-{(x-5)^2\over2}\right)}}\]

Answer 5

yeah...

Answer 6

I don't like it at all!

Answer 7

according to the practice version it seems I take the derivative twice to find the data, it's soo konfusing... All of these questions are weird... :(.

Answer 8

ok this is very ugly so I want you to be \(sure\) that the problem is\[\huge \frac1{\sqrt{2\pi}^{\left(e-{(x-5)^2\over2}\right)}}\]everything is under 1 and all that is correct?

Answer 9

sec adding an attachment.

Answer 10

ugly....

Answer 11

oh that's just terribly written... what the heck is that 2/2 ??? is the exponent 2/2=1
or do they means that whole part is over 2?
somebody messed up I think in typing this out

Answer 12

yeah I was going to say it looks like 2/2 but that would be 1 so ... :(

Answer 13

\[\huge g(x)=\frac1{\sqrt{2\pi}}e^{\frac{-(x-5)^2}2}\]is what it looks like to me I guess...

Answer 14

so you wanna take a stab at that derivative? that's how we start this whole thing....

Answer 15

\[\huge \frac1{\sqrt{2\pi}}{e^{-{x-5^2\over2}}}\] is my guess

Answer 16

I'm scared :(.

Answer 17

there's like an entire train of chain rules :P.

Answer 18

oops i should have looked at the picture

Answer 19

yeah it's chain rule madness

Answer 20

Lets see... I would start off with the e^-(x-5)^2 would be 2(x-5)e^(x-5)^2

Answer 21

unless satellite has some more clever way :D

Answer 22

not so bad at all. ignore the \(\frac{1}{\sqrt{2\pi}}\) as that s a constant and just take the derivative of \(\frac{-(x-5)^2}{2}\)

Answer 23

but there's a chain rule it itself so then there would be a 1 I guess so I guess that doesn't even matter....

Answer 24

yeah true, not really that bad...

Answer 25

then I guess we would need the quotient rule because of that 2?

Answer 26

the derivative if \(-x+5\) or \(5-x\)

Answer 27

or prob not...

Answer 28

no, quotient rule would be if you had a funciton of x in the denominator

Answer 29

*is

Answer 30

oh okay.

Answer 31

the two in the denominator is a constant. leave it

Answer 32

(2(x-5)e^(x-5)^2)/2?

Answer 33

err

Answer 34

(-2(x-5)e^-(x-5)^2)/2 ??

Answer 35

\[\frac{d}{dx}[\frac{-(x-5)^2}{2}=-x+5\] by the power rule

Answer 36

the twos cancel

Answer 37

ah gotcha.

Answer 38

I hate this question... :(

Answer 39

so then it;s (-x+5)e^-(x+5)^2?

Answer 40

well there is still a two in the denominator of the exponent

Answer 41

ah okay. So that orignal 2 came down as well and then they cancelled gotcha.

Answer 42

but before you start hating this question, lets do it in our heads

Answer 43

I'll hate it forever and ever :(. ;P

Answer 44

you have the derivative as some constant times \(-x+5\) times e to the power of something, and you want the critical points.
e to the power of whatever is never zero, so your only job is to set \(-x+5=0\) and solve, which i assert is not that hard!

Answer 45

so it should be 1/sqrt(2pi) -(x+5)e^-(x+5)^2 / 2

Answer 46

x = 5?

Answer 47

stil have no idea :(

Answer 48

and then just solve for y?

Answer 49

yes the critical point is at \(x=5\)

Answer 50

welcome back satellite :P

Answer 51

now to solve for y.

Answer 52

y = that big function right?

Answer 53

where are you stuck on this?

Answer 54

how to find y? is it = to the big nasty eq?

Answer 55

just plug in for x and just go huh?

Answer 56

and I bet I gotta use logs too...

Answer 57

yes, the maximum will be found by pluggin x=5 into:\[\huge g(x)=\frac1{\sqrt{2\pi}}e^{\frac{-(x-5)^2}2}\]

Answer 58

no logs - note that you will end up with \(e^0\) which is just equal to one

Answer 59

do you understand?

Answer 60

wouldn't it be 1/2?

Answer 61

woops nvm.

Answer 62

so we then have 1/(sqrt(2pi) = y?

Answer 63

\[g(5)=\frac{1}{\sqrt{2\pi}}\]

Answer 64

this represents the maximum value for the function g(x)

Answer 65

to find the points of inflection you need to differentiate g(x) twice, equate it to zero, and then solve to find which value(s) of x make it zero

Answer 66

g(x) = 1/sqrt(2pi) -(x+5)e^-(x+5)^2 / 2 ??

Answer 67

???

Answer 68

g(x)'

Answer 69

haha :)

Answer 70

lemme retype that :P.

Answer 71

\[\frac{1}{\sqrt{2\pi}} -(x+5)e^\frac{{-(x+5)^2}}{2}\]

Answer 72

you should get:\[g'(x)=\frac{(x-5)e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}\]

Answer 73

now differentiate this again to get the second derivative

Answer 74

ok

Answer 75

\[\frac{2(x-5)(x-5)e^-\frac{(x-5)^2}{2}}{\sqrt{2\pi}}\] ?

Answer 76

But I'm guessing the 2 in front cancels with the 2 like from before...

Answer 77

it is actually:\[g''(x)=\frac{(x-5)^2e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}-\frac{e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}\]\[\qquad=\frac{((x-5)^2-1)e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}\]\[\qquad=\frac{(x^2-10x+25-1)e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}\]\[\qquad=\frac{(x^2-10x+24)e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}\]\[\qquad=\frac{(x-4)(x-6)e^{-\frac{(x-5)^2}{2}}}{\sqrt{2\pi}}\]

Answer 78

What the poop....

Answer 79

now find what values of x make this zero

Answer 80

so where do we square it, and get the other e business?

Answer 81

you need to refresh yourself on how to differentiate. this is using the chain rule.

Answer 82

yeah I figured, just trying to figure out from what :P.

Answer 83

Well I guess we bring another (x-5) down so that's where the ^2 is? And then we leave the other e^ sfdjdlskfjdslk which causes a - that?

Answer 84

roughly - yes

Answer 85

here, I entered this into wolframalpha for you: http://www.wolframalpha.com/input/?i=d^2%2Fdx^2+%281%2Fsqrt%282*pi%29%29*e^%28-%28x-5%29^2%2F2%29

click "show steps" to see how the derivative is calculated

Answer 86

okies.

Answer 87

what is left in this problem is as follows:

1. find the value(s) of x that make g''(x)=0
2. find the corresponding value(s) of g(x) for each solution in step 1

this will give you your point(s) of inflection.

Answer 88

that damn product rule! :P

Answer 89

would 0 work?

Answer 90

and 12?

Answer 91

144- (10*12[120]) + 24?

Answer 92

or maybe it's 4 and 6 :P

Answer 93

yeah it's 4 and 6, and the y's for both are 0.242 :)

Answer 94

that is the correct answer - x=4 and x=6 are the x-coordinates of the inflection points.
I haven't worked out the y-coordinates but assume you have done it correctly by working out g(4) and g(6).

Answer 95

\((x-5)^2\) would give 1 for both x=4 and x=6, so both y values are indeed the same.