Question

@jim_thompson5910

I have some problems here I wouldn't mind you checking/the 3rd one helping me answer them. That'd be swell :)

Answer 1

Just let me attach a file in a second

Answer 2

#1 is correct

Answer 3

For number 3 the answer choices are:

f(0)=2
f(x) does not equal 2 for all x>=0
f(2) is undefined
lim x approaches 2 f(x)= infinity
lim x approaches infinity f(x)=2

I think It's the 3rd one but I'm not fully convinced in myself.

Answer 4

#2 is a bit off

the integral of e^(-4x) is NOT -4e^(-4x)

Answer 5

Oh darn then what is it?

Answer 6

it's (-1/4)*e^(-4x)

Answer 7

you can check by deriving (-1/4)*e^(-4x)

y = (-1/4)*e^(-4x)

y' = d/dx[ (-1/4)*e^(-4x)]

y' = -4*(-1/4)e^(-4x)

y' = e^(-4x)

Answer 8

Oh right is it because you set u=-4x and du=-4dx so -1/4du=dx?

Answer 9

To integrate e^(-4x), I'd let u = -4x

so du = -4dx ---> dx = -du/4 ---> dx = (-1/4)dx

Answer 10

exactly

Answer 11

So it's 1/4-e^-4/4

Answer 12

one sec

Answer 13

If you meant

\[\Large \frac{1}{4} - \frac{1}{4}e^{-4}\] then you are correct

Answer 14

Well it's the same thing as \[\frac{ 1 }{ 4 }-\frac{ e ^{-4} }{ 4} \]

Answer 15

correct

Answer 16

Cool and for the 3rd one? I think it's the 3rd answer but not fully sure

Answer 17

I see the answer choices, but does it say what the function f is?

Answer 18

oh nvm, we don't need the function lol

Answer 19

f(2) may or may not be defined, it's hard to say

but all we know is that the horizontal asymptote is y = 2

Answer 20

so if you the horizontal asymptote is y = 2, what can you say for sure about f(x)?

Answer 21

Uhmmm I think that f(x) cannot equal 2 for all x>=0?

Answer 22

WAIT no no no

Answer 23

a function/graph can cross the horizontal asymptote

Answer 24

It's an asymptote so the limit as x approaches 2 f(x)= infinity

Answer 25

you just described the vertical asymptote at x = 2

Answer 26

Oh darn I guess I wasn't visualizing (I was bad at the limits chapter unfortunately)

So is it f(0)=2 because it's a horizontal line and they have a slope of 0? I'm not sure to be honest.

Answer 27

you were on the right track when you brought up the limit

Answer 28

Oh I couldn't visualize the other limit so I just guessed :P

Answer 29

as x approaches infinity, then f(x) will approach 2

it could cross over the asymptote (as many times as you want), but overall, as x gets really huge, f(x) will settle on 2

Answer 30

and we write that as

\[\Large \lim_{x\to \infty} f(x) = 2\]

Answer 31

Oh right is that the thing like this

Answer 32

yeah that's one of many ways to visualize it

Answer 33

Oh alright I understand now

Answer 34

that's good

Answer 35

Can you check the first two and help me with the 3rd? I suck at limits

And in the second one it's rad2/2-1 I just forgot to rewrite the 2 in the denominator because it looked messy

Answer 36

#4 is correct. Nice job

Answer 37

for #5, the integral of sin(x) is -cos(x)

not just cos(x)

Answer 38

this is because if y = cos(x), then dy/dx = -sin(x)

so we get close to what we want, we're just off by a sign

if we have y = -cos(x), then dy/dx = -(-sin(x)) = sin(x)

Answer 39

Oh right

Answer 40

So -rad2/2+1

Answer 41

good

Answer 42

And for the 3rd I think I have to do long division then solve for x but i dont remember how to do long division for polynomials lol

Answer 43

you could do long division, but it would take a very long time and there's a potential for error

Answer 44

for #6, you can divide everything by x^3 to go from

\[\Large \frac{x^3 - 2x^2+3x-4}{4x^3 - 3x^2+2x-1}\]

to

\[\Large \frac{\frac{x^3}{x^3} - \frac{2x^2}{x^3}+\frac{3x}{x^3}-\frac{4}{x^3}}{\frac{4x^3}{x^3} - \frac{3x^2}{x^3}+\frac{2x}{x^3}-\frac{1}{x^3}}\]

and that turns into

\[\Large \frac{1 - \frac{2}{x}+\frac{3}{x^2}-\frac{4}{x^3}}{4 - \frac{3}{x}+\frac{2}{x^2}-\frac{1}{x^3}}\]

Answer 45

Why would I divide everything by x^3?

That looks really unfamiliar to me he always has us to long division and I never remember lol

Answer 46

like I said, you could do long division, but be cause the denominator is so large, it'd take a while

Answer 47

I'm dividing by x^3 since it's the largest term with the largest exponent

Answer 48

doing that will turn each term into a mini fraction (within the fraction)

Answer 49

each mini fraction will then go to 0 as x --> infinity

Answer 50

Okay and then what would you do after that?

Answer 51

so it effectively eliminates a lot of the terms

Answer 52

\[\Large \frac{1 - \frac{2}{x}+\frac{3}{x^2}-\frac{4}{x^3}}{4 - \frac{3}{x}+\frac{2}{x^2}-\frac{1}{x^3}}\]

turns into

\[\Large \frac{1 - 0 + 0 - 0}{4 - 0 + 0 - 0}\]

when x approaches infinity

Answer 53

What how?

Answer 54

for instance

\[\Large \lim_{x \to \infty} \left(\frac{2}{x}\right) = 0\]

Answer 55

each mini fraction will have the same limit...0

Answer 56

Oh right then I don't even have to do the divide by x^3, I can just plug in 0 and get 4 can't I?

So wait idk lol

Answer 57

not quite

Answer 58

remember x is approaching infinity, not 0