Question

@jim_thompson5910

http://cyh.leeschools.net/UserContent/Documents/AP%20CalcBC%20SumAssign%2014-15.pdf

Answer 1

which one are we on now?

Answer 2

81 i think?

Answer 3

81

And I believe I need to find the critical numbers and make myself a chart

Answer 4

correct

Answer 5

and you derive before all that, but I'm sure you had that in mind

Answer 6

actually nvm, it's already derived for you

Answer 7

It's already derived :P

Answer 8

@marissalovescats wait until you see @jim_thompson5910 invoice....

Answer 9

yeah caught myself too late lol

Answer 10

What? @satellite73

Answer 11

just kidding

Answer 12

he's making a joke on how I'd charge, but this site is all free lol

Answer 13

you don't get paid?
i get \($10\) per answer, \(\$20\) if they are correct

Answer 14

Oh right, that's what I thought haha.

Yeah next year I'll owe my grade in AP Calc BC, AP Statistics, and possible AP Physics to him.

#jimhelpsmarissagettocollege. Let's get it trending.

Answer 15

"Helper on free website OpenStudy helps local girl climb in class rank from #6 to hopefully #3 or above"

Exactly.

Answer 16

yeah offsite though

Answer 17

here it's all free, so don't worry about that

Answer 18

anyways, what critical values did you get?

Answer 19

Okay so # 81 :P

Not sure about finding the critical number to be honest, can't take a square root of a negative number

Answer 20

My pay to you will be idk I'll send you copies of my transcript and copies of my acceptance letters and the scholarships I receive.

Answer 21

we have sin(x^2 + 1) = 0

how do we get rid of that "sin" and move it to the other side so to speak

Answer 22

Question:

What about the part of the question "on the interval 2 to 4" What's it want? Lol

Answer 23

I'm not sure because it's being multiplied with the (x^2+1)

Answer 24

it's not being multiplied with sin

sin is a function it's the sine function

Answer 25

Marissa does math all day and then can't remember her Calculus basics. Woo :P

Answer 26

we would apply the arcsine to both sides to get

sin(x^2 + 1) = 0

arcsin(sin(x^2 + 1)) = arcsin(0)

x^2 + 1 = arcsin(0)

Answer 27

Which would just be x^2+1=0 no?

Answer 28

Wow 2 medals already, who would even medal you it's not like you actualllly help me :P

Answer 29

that's one equation yes, since sin(0) = 0

but because sin(pi) = 0 as well, this means we go from this

x^2 + 1 = arcsin(0)

to this

x^2 + 1 = 0 or x^2 + 1 = pi

Answer 30

Oh interesting

Answer 31

use the unit circle to see that sin(0) = 0 and sin(pi) = 0

Answer 32

Yeah that much I know lol

Answer 33

technically arcsin(0) = 0 but there are two solutions to sin(x) = 0, which are x = 0, x = pi

Answer 34

Yeah

So x^2+1=0 and x^2+1=pi

But how do I solve for x in those? Haha

Answer 35

you don't have to solve really

you just have to count the number of solutions of each and then add up those counts

Answer 36

actually nvm

Answer 37

yeah you have to solve

Answer 38

so subtract 1 from each side

then take the square root of both sides (don't forget about the plus/minus)

Answer 39

But you cant take the square root of a negative number?

Answer 40

that is true, so you ignore any non-real solutions

Answer 41

x^2=-1 and
x^2=-1+pi?

Answer 42

yep

Answer 43

So.. x=+-1
and x=+-sqrt(1+pi)

?

Answer 44

x^2=-1 has no real solutions

Answer 45

x^2=-1 --> x = i, x = -i

Answer 46

Alrighty..

What about the other one?

Answer 47

x=+-sqrt(-1+pi) should be, so we have

x=sqrt(-1+pi) or x=-sqrt(-1+pi)

Answer 48

Alrighty so now what?

Answer 49

find the approximations to x=sqrt(-1+pi) or x=-sqrt(-1+pi)

Answer 50

Oh okay

Answer 51

1.46 and -1.46

Answer 52

is that in the range 2 < x < 4 ?

Answer 53

Nope

Answer 54

what about if you add 2pi to each (2pi = 2*3.14 = 6.28 roughly)

Answer 55

No

Answer 56

keep in mind that adding 2pi just finds coterminal angles

Answer 57

Isn't relative extrema mins and maxs?

Answer 58

one sec, let me think

Answer 59

Relative extrema are the relative minimums and maximums

Answer 60

yes that's true

Answer 61

Sooo what do we do to solve this problem? Lol

Answer 62

oh I'm doing it all wrong...ugh lol

Answer 63

I kinda had a feeling we were :P

Answer 64

I was like none of this looks familiar

Answer 65

it should be like this

sin(x^2 + 1) = 0

x^2 + 1 = arcsin(0)

x^2 + 1 = pi*n ... n is any integer

x^2 = -1 + pi*n

x = +-sqrt( -1 + pi*n ) ... again, n is any integer

Answer 66

Honestly I just thought the critical numbers where 0 and pi

I thought to divide by x^2+1 and have sin=0

Answer 67

sin=0 doesn't make sense though

sin(x) = 0 makes more sense

Answer 68

We've never done anything with imaginary numbers either

And yeah that's true

Answer 69

to get your roots, you need to find the approximation to x = +-sqrt( -1 + pi*n ) for various values of n

Answer 70

Whatever that means haha

Answer 71

when n = 0, you have

x = +-sqrt( -1 + pi*n )

x = +-sqrt( -1 + pi*0 )

x = +-sqrt( -1 )

so we don't have any real roots here

Answer 72

when n = 1

x = +-sqrt( -1 + pi*n )

x = +-sqrt( -1 + pi*1 )

x = 1.46341 or x = -1.46341

Answer 73

those values are real, but not in the range 2 < x < 4

Answer 74

you keep going until you find the roots that all lie in 2 < x < 4

Answer 75

what do we use for n?

Answer 76

any integer you want

Answer 77

that generates all of the solutions to sin(x^2 + 1) = 0

Answer 78

Uhm okay

4?

Answer 79

that's the correct final answer

hopefully you're finding all of the right roots

Answer 80

What?

Answer 81

oh nvm, jumping the gun lol

Answer 82

x=2.9 and -2.9

Answer 83

yeah use n = 1, n = 2, n = 3, n = 4, etc

Answer 84

I'm so confused.

Answer 85

since 2.29 is in range of 2 < x < 4, that's one root

Answer 86

we're just finding the roots to sin(x^2 + 1) = 0

Answer 87

the roots are all of the form x = +-sqrt( -1 + pi*n )

Answer 88

But numbers are in the range too if we use 3,4, or 5

Answer 89

correct

Answer 90

there are 4 roots total that are in the range 2 < x < 4

Answer 91

So are allllll of those going to be roots?

Answer 92

I mean critical numbers

Answer 93

there are infinitely many roots BUT there are only 4 roots that are in the range 2 < x < 4

Answer 94

How

Answer 95

when n = 2

x = +-sqrt( -1 + pi*n )

x = +-sqrt( -1 + pi*2 )

x = +-2.2985 ... one value is in range

------------------------

when n = 3

x = +-sqrt( -1 + pi*n )

x = +-sqrt( -1 + pi*3 )

x = +-2.9025 ... one is in range

------------------------

when n = 4

x = +-sqrt( -1 + pi*n )

x = +-sqrt( -1 + pi*4 )

x = +-3.4009 ... one is in range

------------------------

when n = 5

x = +-sqrt( -1 + pi*n )

x = +-sqrt( -1 + pi*5 )

x = +-3.835096 ... one is in range

------------------------

Answer 96

if you tried n = 6, you'd find that you're out of range of 2 < x < 4

Answer 97

Right so there are 3 relative extrema?