Question

How close to −3 do we have to take x for
the inequality
1/(x + 3)^2 > 700
to hold?
1. within at least 0.097
2. within at least 0.057
3. within at least 0.037
4. within at least 0.077
5. within at least 0.017

Answer 1

x has to be within
\[\sqrt{\frac{1}{700}}\] of -3 ,whatever that number is

Answer 2

.037 is the nearest

Answer 3

need the method?

Answer 4

oh so the -3 has nothing t do with the problem? it's basically seeing if it falls on that range?

Answer 5

oh it has something all right
flip the inequality to get
\[(x+3)^2<\frac{1}{700}\] take the square root get
\[x+3<\frac{1}{\sqrt{700}}\]

Answer 6

well that is a mistake sorry
lets do it right

Answer 7

\[(x+3)^2<\frac{1}{700}\]
\[-\frac{1}{\sqrt{700}}<x+3<\frac{1}{\sqrt{700}}\]

Answer 8

or
\[|x+3|<\frac{1}{\sqrt{700}}\]

Answer 9

which is the math way to say x is within .037 of -3

Answer 10

gotcha!

Answer 11

go ahead and post i will take a look

Answer 12

im having trouble with 5,6,7,11,12,13,14 :/

Answer 13

that is a lot of troubles! i will do 5 first. the function
\[g(x)=\frac{2x-5}{|x-7|}\] has a vertical asymptote at x = 7. i am not sure why it says it is a "jump" discontinuity, but never mind.

Answer 14

as you approach from the right it goes to positive infinity. f(7) = 0 but it is positive on the right. so that limit will be positive infinity

Answer 15

\[f(x)=\frac{x^2-x-20}{x^2-7x+10}\]

Answer 16

its not positive infinity :/

Answer 17

hold on let me reread the problem, but let me finish #6

Answer 18

\[f(x)=\frac{(x-5)(x+4)}{(x-5)(x-2)}=\frac{x+4}{x-2}\]

Answer 19

you can't get rid of the discontinuity at 2

Answer 20

if you replace x by 5 you get
\[\frac{9}{3}\] which is not
\[\frac{10}{3}\] so the function is not continuous at 5 either

Answer 21

#7
C holds for sure, because the function is continuous on [-2,2] and negative at -2 , positive at 2, so it must cross the x axis somewhere
B is silly, you have no idea what the function looks like.
\[f(-2)+(-2)=-1, f(2)+2=1\] so
\[f(x)+x\] must also cross the x axis

Answer 22

so A and C

Answer 23

if it is not positive infinity what is it?

Answer 24

idk:/ im thinking it doesnt exist

Answer 25

hmmm i don't know why. hold on

Answer 26

that is the one with the graph right?

Answer 27

oh crap i was wrong. sorry. does not exist is right i believe my mistake

Answer 28

#11
at 1:00 accelerating because slope of tangent lines are increasing. as for first part in 10 minutes he went 40 km

Answer 29

so that is exactly 120 km/hr average, not less
these are tricky!

Answer 30

where did you get these?

Answer 31

Lol my teacher sends them to me for a review.

Answer 32

12 is actually easiest of all

Answer 33

is your teacher uri treismann?

Answer 34

yess

Answer 35

lol

Answer 36

omg number 5 isnt that the limit doesnt exist lol

Answer 37

wtf

Answer 38

what is it???

Answer 39

omg it isnt -7..

Answer 40

i have to submit my answer and it tells me if its wrong or not. if it is, i gotta try again

Answer 41

let me post this damned problem and see if we get a response. but first lets do 12

Answer 42

ready?

Answer 43

the derivative is the slope of the tangent line. if you look at the picture you will see that at -2 the function is going down, so
\[g'(2)<0\]

Answer 44

it was 9. and 11 was B only

Answer 45

at 0, 1 and 3 it is increasing , so we know that the order begins
\[g'(2),0,...\]

Answer 46

yes we knew 11 was B only. maybe my answer was not clear. it was exactly 120 km/hr , not less than

Answer 47

9???!!!

Answer 48

hold on let me post it. but 12 should be easy to finish. the derivative is the slope of the tangent line, so arrange so that slopes go from smallest to biggest

Answer 49

#6 is correct for problem 12

Answer 50

i just need number 13 :D