I need AP Calculus help!!

- anonymous

I need AP Calculus help!!

- Stacey Warren - Expert brainly.com

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- chestercat

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- anonymous

i will post the assignment soon!

- anonymous

Help with all plz!

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- anonymous

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## More answers

- anonymous

hi

- anonymous

the circled questions btw

- anonymous

for the first one, its just 8 thats posted?

- anonymous

i mean circled

- anonymous

its numbers 10, 14, 16, 18, 22

- anonymous

oh i was looking at the first one

- anonymous

kk

- anonymous

oh man, that second one's really hard to read lol so lets do the first one

- anonymous

kk

- anonymous

k so for 8 what is x approaching? i cant see

- anonymous

we dont have to do number 8 lol

- anonymous

:\ it looked circled to me lol

- anonymous

its only numbers 10, 14, 16, 18, 22

- anonymous

do you not have to do the problem in your first attachment?

- anonymous

oh yeah duh i forgot i do lolz

- anonymous

haha ok cool well i basically cant read any of them cuz they're blurry so can you just type out one that you need help with?

- anonymous

kk

- anonymous

hang on

- anonymous

number 8 its as x goes to infinity

- anonymous

so it's as x-> infinity of (2x^2)/(x^2 +3) ? is that right?

- anonymous

2x^5/ x^5+3

- anonymous

oh ok.

- anonymous

have y'all talked about L'Hospital's Rule yet?

- anonymous

i dont think so

- anonymous

k.

- anonymous

what is the L'hospitals rule? can u plz explain

- anonymous

i don't think you're supposed to use it if you haven't covered it yet. you don't need it anyway.

- anonymous

so you start by considering the terms with the highest exponents

- anonymous

if I remember correctly, since x is approaching infinity and it's a rational function, you can rewrite it only including the x's with the greatest exponents

- anonymous

so what would you have if you did that?

- anonymous

hang on im kinda confused

- anonymous

that's ok. so what i'm saying is your next step is to rewrite the limit excluding the +3 in the denominator

- anonymous

to exclude the +3 in the denominator how do u do tht?

- anonymous

you can do this because when x goes to infinity, 2x^5 and x^5 approach infinity and the 3 basically doesn't change the fraction anymore

- anonymous

because if you imagine really high numbers, say x=100, x^5 would be 10000000000 and 2x^5 would be 20000000000 and the +3 in the denominator makes less and less of a difference in the fraction. (10000000000/20000000003 is really close to 10000000000/20000000000)

- anonymous

oh ok

- anonymous

in other words, the fraction approaches 1/2

- anonymous

im sorry, it approaches 2

- anonymous

because 2x^5 is in the numerator

- anonymous

oh ok so ur limit is basically 2 then right?

- anonymous

yup

- anonymous

oh ok now i get number 8

- anonymous

so in questions like that, where x is approaching infinity, you can disregard any terms that arent the variables with the highest exponent. like if you had an x^4 being added/subtracted somewhere in the fraction, you could disregard that, too.

- anonymous

why can u?

- anonymous

for the same reason, because as x approaches infinity, they change the fraction less and less, even though it has x in it....those terms approach infinity slower than terms of higher powers.

- anonymous

im kinda confused on tht part

- anonymous

it's kind of counter-intuitive at first

- anonymous

oh ok

- anonymous

as x gets greater, the difference between x^5 and x^4 also gets greater. if x is 10, then x^5 is 100000 and x^4 is 10000, x^5 is 10 times greater than x^4. but when x= 100, x^5 becomes 100 times bigger than x^4, and so on, until x^5 is infinitely bigger than x^4 and you can disregard x^4.

- anonymous

oh ok now i get it

- anonymous

:) cool

- anonymous

can u help me with the continuity limits and one-sided ones those ones are the ones that r confusing

- anonymous

i can help w one more then i need to get my own work done haha

- anonymous

kk

- anonymous

so i guess pick one for us to do

- anonymous

kk

- anonymous

i think i get the rest on my THQ but i need help with the continuity ones

- anonymous

k

- anonymous

let me post the problem

- anonymous

kk

- anonymous

\[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 } \]

- anonymous

so first notice that when x = 4, f(x) = 0/0, which is undefined.

- anonymous

yeah

- anonymous

and since x is approaching 4 from the left (less than 4, getting greater) both the numerator and denominator are going to be negative...do u get that?

- anonymous

can u explain it a bit more im still a bit confused

- anonymous

sure. so the way i think of it, is to basically imagine x as some number a little less than 4.. 3.something. when you plug that number into the denominator, 3.something - 4 is always going to be negative. (anything less than 4) - 4 will be negative, right? (3-4=-1, 3.5-4=-0.5, 3.99-4 = -0.01)

- anonymous

oh yeah so its like it keeps on approaching 4 right?

- anonymous

and for the numerator, you can imagine that since the square root of 4 is 2, the square root of anything less than 4 will be less than 2... when you subtract 2 from that number, the result is negative

- anonymous

ya it keeps approaching 4

- anonymous

but from the less-than-4 side because it has the minus sign (x->4-)

- anonymous

basically the limit for it will negative right?

- anonymous

well the numerator and denominator are both negative, right?

- anonymous

so the fraction comes out positive

- anonymous

oh ok

- anonymous

why r limits soo confusing :3

- anonymous

lol i know. wanna see my limit problem that no one will answer?

- anonymous

Limit as x->(pi/2) of [tanx+1/(x-pi/2)]?

- anonymous

thats my question that no one will answer :(

- anonymous

i had it on a quiz today

- anonymous

can u do the substitution method for tht one

- anonymous

i think

- anonymous

idk

- anonymous

no because tan of pi/2 is undefined, it goes to neg infinity and infinity from the right and left....

- anonymous

oh yeah ur right oops lol

- anonymous

ya...it's brutal.

- anonymous

anyway

- anonymous

but im still kinda confused on the continuity limits and one-sided ones

- anonymous

oh boy satellite's here.... it's about to get real xD

- anonymous

i just realized that your problem is tricky. you need to factor your x-4 into (sqrt(x)-2)(sqrt(x)+2) then cancel out the numerator

- anonymous

\[\lim_{ x\to\frac{\pi}{2}}\frac{\tan(x)+1}{x-\frac{\pi}{2}}\]?

- anonymous

oooh well that's my problem but not hers

- anonymous

mines posted

- anonymous

oh ok

- anonymous

somewhere down the list!

- anonymous

@jldstuff393 how do u do my problem kinda confuzzled

- anonymous

i know i know. u remember difference of squares? like (x^2 - 4) = (x-2)(x+2)?

- anonymous

is this the problem
\[\lim_{x \rightarrow 4-} \frac{ \sqrt{x}-2 }{ x-4 }\]?

- anonymous

oh yeah i remember different of squares

- anonymous

yeah @satellite73 thts my problem

- anonymous

so unless satellite has a less tricky way to solve it, you can factor (x-4) as a difference of squares... (sqrt(x)-2)(sqrt(x)+2)

- anonymous

does that make sense?

- anonymous

the gimmick here is to "rationalize the numerator" by multiplying by the conjugate of \(\sqrt{x}-2\) which is \(\sqrt{x}+2\)

- anonymous

cuz when u have that, you can cancel out the (sqrt(x)-2) in the numerator and denominator

- anonymous

so u multpily the top and bottom by \[\sqrt{x} +2 \]

- anonymous

this works because \((a+b)(a-b)=a^2-b^2\) and so \((\sqrt{x}-2)(\sqrt{x}+2)=x-4\)

- anonymous

that cancels with the \(x-4\) in the denominator, which is what you want, you want to cancel the zero

- anonymous

and x-4 can cancel in denominator

- anonymous

oh ok!

- anonymous

so ur limit is one right?

- anonymous

leaves you with
\[\lim_{x \rightarrow 4} \frac{ \sqrt{x}-2 }{ x-4 }=\lim_{x \rightarrow 4} \frac{1 }{ \sqrt{x}+2}\]\]

- anonymous

oh ok i get

- anonymous

no limit is not one, now you have to replace \(x\) by 4

- anonymous

once u replace x with 4 u get 1/ 4 right?

- anonymous

yup

- anonymous

whenever you get \(\frac{0}{0}\) you need some gimmick to factor and cancel to get rid of the zeros

- anonymous

satellite would you mind helping me after you're done here? O.o good luck guys!

- anonymous

oh ok

- anonymous

sometime sit is easy
sometimes it is a pain.
but you get used to it

- anonymous

@satellite73 i have some more questions!

- anonymous

@jldstuff393 sure if i can
what is the question

- anonymous

@abannavong go ahead and post them, i will look

- anonymous

i'll repost it, thanks!

- anonymous

kk! thanks @satellite73

- anonymous

14, 16, and 18

- anonymous

i can't really read it, but for 16 replace \(x\) by 2 in both formulas
if you get the same number, that is the limit
if you get different numbers, there is no limit

- anonymous

number 14 its as deltax goes to 0 with the little plus sign

- anonymous

i can't read 18 either
is it \(\lim_{ x\to 1^+}\) or \(\lim_{x\to 1^-}\)?

- anonymous

x-> 1+

- anonymous

then replace \(x\) by 1 in the second formula
you get \(1-1=0\)

- anonymous

oh ok

- anonymous

can u help with 14 also

- anonymous

its confusing

- anonymous

sure, but i can't really read it
can you post in a new question? you do not need to write \(\Delta x\) you can just write \(h\) instead

- anonymous

it is actually very straight forward
we will do a little algebra and that is all

- anonymous

kk hang on dinner quickly

- anonymous

\[(x+h)^2+x+h-(x^2+x)=x^2+2xh+h^2+x+h-x^2-x\]
\[\cancel{x^2}+2xh+h^2+\cancel{x}+h-\cancel{x^2}-\cancel{x}=2xh+h+h^2\]
divide by \(h\) and get \(2x+1+h\)
let \(h\to 0\) be replacing \(h\) by 0 and you get \(2x+1\)

- anonymous

oh ok! @satellite73 im back from dinner

- anonymous

that was a quick meal, you should take your time
i wrote the solution above, but i used \(h\) instead of \(\Delta x\)
it is mostly just algebra, to get rid of the factor of \(\Delta x\) in the denominator, then you replace it by 0

- anonymous

i did eat dinner

- anonymous

im eating and doing work lolz

- anonymous

you are either eating late or are on the left coast

- anonymous

im on the west coast lolz

- anonymous

after you finish eating, read my solution above
i hope all the steps are clear
they are almost all algebra steps, only at the end do you replace \(h\) by 0

- anonymous

kk

- anonymous

oh ok its jst basic algebra right? so u foil the (x+h)^2 right?

- anonymous

yes exactly
it is \((x+h)(x+h)=x^2+2xh+h^2\)

- anonymous

oh ok and then from there u can combine the like terms and get 2xh+h+h^2

- anonymous

yes,

- anonymous

that is all in the numerator
you get
\[\frac{2xh+h+h^2}{h}\]

- anonymous

and then can u factor out the h on the numerator

- anonymous

yes factor and cancel or divide each term by \(h\) it is the same thing

- anonymous

oh ok and then u get 2x+1

- anonymous

yes

- anonymous

oh ok! i get tht now

- anonymous

whew that was exhausting

- anonymous

\[\lim_{x \rightarrow2 ^{+}}(2x-\left[ \left| x \right| \right]\] help im confused with this one also

- anonymous

is that an absolute value sign?

- anonymous

or is that the floor function?

- anonymous

its like a bracket and then two straight lines inside

- anonymous

do you know what that function is called?

- anonymous

ughhh i dont remember what it was called

- anonymous

it is either called "floor" or "greatest integer"

- anonymous

is it kinda like a piecewise function or something like tht

- anonymous

oh yeah i remember!

- anonymous

so if \(x>2\) but close to 2, then \([x]=2\)

- anonymous

so you have
\[\lim_{x\to 2}2x-2=4-2=2\]

- anonymous

to be more precise
\[\lim_{x\to 2^+}2x-2=4-2=2\]

- anonymous

oh ok! at first it seemed confusing but oh ok now i get it!

- anonymous

whereas
\[\lim_{x\to 2^-}2x-2=4-1=3\] because if \(x<2\) but close to 2, then \([x]=1\)

- anonymous

oh ok! tht seems soo much easier now!

- anonymous

yes if you know what it means it is not hard
here is a nice picture of \(y=2x-[x]\)
http://www.wolframalpha.com/input/?i=2x-floor%28x%29

- anonymous

oh ok! i get it! i have another question again hang on

- anonymous

so it says find the x-values(if any) at which f is not continuous. which of the discontinuities are removable.
f(x)= 1/ x^2+1

- anonymous

\[f(x)=\frac{1}{x^2+1}\] has no discontinuities, since it is a rational function and the denominator is never 0
since \(x^2\geq 0\) for any \(x\) you can see that \(x^2+1\geq 1\) no matter what \(x\) is, so this function has no discontinuities. as the denominator is never zero

- anonymous

oh ok thank you @satellite73

- anonymous

you are welcome
enough math, study something else
or do something more constructive

- anonymous

pretty much tht was my last problem! time to find a prayer for theology class! thank you @satellite73 and study for english test tomorrow on grapes of wrath

- anonymous

good luck with the joads

- anonymous

lolz thanks i have to find the intercalary chapters to write about :3 i dont even know what those r

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