Question

calc help @angel

Answer 1

wrong person oops

Answer 2

@Shadow

Answer 3

1. you need to know what the signs of the derivatives mean

> first derivative positive = function is increasing
> second derivative positive = function is concave up

> first derivative negative = function is decreasing
> second derivative negative = function is concave down

Answer 4

2. to find any extrema within that interval, take the derivative and set it equal to 0, solve for x. any x-values between -2 and 1 should be considered, and evaluted to see which one gives you the absolute max when plugged back into the original function (not the derivative, the original function)

Answer 5

3. you already have the derivative so solve for f'(x) = 0 and add up all the x-values, being mindful of signs

Answer 6

4. you will have to find the range of t-values where the first derivative is pos. and the second deriv. is negative

Answer 7

is 1 c then? how would I sow my work for that question

Answer 8

it's a definition question so you could just write "definition of derivatives and concavity"

Answer 9

"first derivative is negative so the function must be decreasing since first derivative gives rate of change" "second derivative is negative so the function must be decreasing according to the definition of concavity"

Answer 10

I am right yay!

Answer 11

2
x=-1

Answer 12

good start but now you have to plug that into the function to get the y-value

Answer 13

13

Answer 14

check your calculations again, you probably missed a negative sign somewhere

-2(-1)^2 -4(-1) + 7 =

Answer 15

9

Answer 16

is that the answer?

Answer 17

yes

Answer 18

3

Answer 19

ah

for 3) I forgot to mention, critical points can also happen when the derivative is undefined

so since you have x - 5 in the denominator, x = 5 is another critical point

so 3+ 5 = 8 = your solution

Answer 20

for 4 do I just plug in values?

Answer 21

after takinf the derivative?

Answer 22

you'd find the critical points first then yes, you can plug in values to see what the signs are

Answer 23

but I do take the derivative right?

Answer 24

and =0?

Answer 25

yes, first and second

Answer 26

yes, set them equal to 0

Answer 27

I'm working hold on lol and then 5 more after this and then no more calc for the semester that you have to help me with

Answer 28

wait the first and second derivative are the same?

Answer 29

without the x?

Answer 30

first derivative is -1/3 e^(-x/3) (x^2 - 6 x + 3)

take the derivative again to get the second derivative

Answer 31

oh it's 0 and first is 2x/3sqrt e

Answer 32

how did you get first answer have to show work I'll work on second derivative

Answer 33

product rule

Answer 34

this?

Answer 35

(t^2+3)e^(-t/3)


(t^2+3)e^(-t/3) * (-1/3) + (2t)e^(-t/3)

factor out (-1/3)e^(-t/3)

(-1/3)e^(-t/3) * (t^2 + 3 - 6t)

Answer 36

for first derivative?

Answer 37

yes.

Answer 38

is my answer for second derivatve correct?

Answer 39

yes

Answer 40

so now set to 0?

Answer 41

yes

Answer 42

if you're using software, to save some time, set first derivative > 0 and second derivative < 0

Answer 43

you will end up with two x-intervals, then find the appropriate x-intervals that are satisfied by both inequalities

Answer 44

It's all written on paper unfortunately

Answer 45

let me what I can do

Answer 46

hm. I wouldn't really know how to solve these by hand easily

Answer 47

hmm it won't let me what site do I use?

Answer 48

oh, duh

(-1/3)e^(-t/3) * (t^2 + 3 - 6t)

for the first derivative you just have to consider this part

(t^2 + 3 - 6t)

since e^(t/3) will never be 0

so set (t^2 + 3 - 6t) = 0 and use the quadratic formula to find the zero's

Answer 49

3+-sqrt6

Answer 50

good

so you have three intervals

x < 3 - sqrt(6)
3 - sqrt(6) < x < 3 + sqrt(6)
and
x > 3 + sqrt(6)

you can just start plugging in x values to see where it's positive

Answer 51

as a hint, just pick something absurd like -100 and 100 for the outer intervals x < 3 - sqrt(6) and x > 3 + sqrt(6)


and 3 for the inside interval 3 - sqrt(6) < x < 3 + sqrt(6)

Answer 52

keep in mind you don't have to come up with an exact value, just whether the result is positive or negative

Answer 53

.55 5.44

Answer 54

good, so you know the first derivative is positive between .55 and 5.44

Answer 55

now repeat the process to find where the second derivative is negative

Answer 56

the whole thing or part because there isn't a solution fot the entire second derivative

Answer 57

there should be a solution

try factoring it

Answer 58

1/9 e^(-t/3) (t^2 - 12 t + 21) = 0

solve for t then use it to find where the second derivative is negative

Answer 59

still no solution that's weird

Answer 60

like before, we know that 1/9 e^(-t/3) can never equal 0 due to being an exponential function

use the quadratic function on (t^2 - 12 t + 21) = 0

Answer 61

*quadratic formula

Answer 62

-t^2-21/12

Answer 63

-b +/- sqrt(b^2 - 4ac) / (2a)

apply this logic to (t^2 - 12 t + 21) = 0

Answer 64

6+-sqrt15

Answer 65

yup, so once again you have three intervals

x > 6-sqrt(15)
6 - sqrt(15) < x < 6 + sqrt(15)
x > 6 + sqrt(15)

see which of these intervals makes the second derivative negative

Answer 66

2.13 9.87

Answer 67

so b or c

Answer 68

so what we have so far

.55 < x < 5.44
2.13 <x < 9.87
which answer choice satisfies both of these intervals?

Answer 69

c

Answer 70

.55 < x < 5.44

so x < 5.44

Answer 71

last 5 shouldn't be too hard

Answer 72

the answer to 4 is not c, check again and make sure BOTH intervals are satisfied

Answer 73

well 5.44 is actually 5.45 rounded

Answer 74

.55 < x < 5.44
2.13 <x < 9.87


so x must be greater than 2.13 or less than 5.44 for BOTH to be satisfied

Answer 75

anything less than 2.13 violates the second interval and anything greater than 5.45 violates the second interval

Answer 76

a?

Answer 77

good

Answer 78

number 5

Answer 79

uh i'd have to think on this one for a bit

Answer 80

your a and b values are 0 and 9 respectively

so find c where

f'(c) = [ f(9) - f(0) ] / (9 - 0)

Answer 81

so take derivative of function and plug in 9 and 0 ?

Answer 82

if you notice [ f(9) - f(0) ] / (9 - 0) doesn't involve the derivative

so calculate [ f(9) - f(0) ] / (9 - 0) using the regular function, not the derivative

Answer 83

then once you have that value you can calculate the derivative, plug in c, and solve for c where f'(c) = [ f(9) - f(0) ] / (9 - 0)

Answer 84

f(x) = 8x/7 + sqrt(x)

x = 0 ----> f(0) = 0
x = 9 ----> f(9) = (8*9)/7 + 3

[ f(9) - f(0) ] / (9 - 0) = [ (8*9)/7 + 3 ] / 9

= (8/7) + (1/3)

then solve f'(c) = (8/7) + (1/3) for c

Answer 85

is the answer false?

Answer 86

I got 1.48?

Answer 87

so the derivative ends up being (8/7) + (1/2)x^(-1/2) right?

(8/7) + (1/2)x^(-1/2) = (8/7) + (1/3)

cross out 8/7 on both sides

(1/2)x^(-1/2) = 1/3

2x^(1/2) = 1/3

x^(1/2) = 1/6

so yeah it's not 4.14 so the answer is false

Answer 88

anyway for the arrow question I believe you'd take the derivative of tan (theta) wrt

Answer 89

so tan 36 is .73

Answer 90

good keep going

Answer 91

is b 230?

Answer 92

uh hold on

Answer 93

tan(36) = b/a
tan(36)*a = b

tan(36) * da/dt = db/dt

so no, 230 is db/dt, solve for da/dt

Answer 94

after you have da/dt I think you can solve for dc/dt

Answer 95

how do I find a?

Answer 96

tan(36) * da/dt = db/dt
we defined db/dt as 230

so tan(36) * (da/dt) = 230 solve for da/dt

Answer 97

315.07

Answer 98

check your rounding again, 230/tan(36) is a bit closer to 316.57

Answer 99

after that I believe it's just the pythagorean theorem 316.57^2 + 230^2 = (dc/dt)^2

Answer 100

76558.28?

Answer 101

square root of that

Answer 102

anyway I really can't spend all day on this

for 3 it might not necessarily be an inflection point if there's no sign change

Answer 103

276.69

Answer 104

just 3 more questions I don't believe they are too hard

Answer 105

so 8 is false for inflection because of no sign of change?

Answer 106

for 4. take the second derivative, plug in x = 1 and y =12 to solve for a, then take the first derivative and solve for b, then once you have both a and b, solve for the slope at the inflection point (where the second derivative is set to 0) and see which function has the equivalent slope

Answer 107

don't really know 5 sorry

Answer 108

so 8 is correct right and first for 9 is 3ax^2+2bx and second is 6ax+2b

Answer 109

@Vocaloid

Answer 110

@mhchen

Answer 111

could you help 1 1/2 questions left

Answer 112

@Hero

Answer 113

@Vocaloid is 9 c?

Answer 114

or 4 c

Answer 115

Post your question as a separate question.