Question

calc help @mhchen

Answer 1

For the first question. Horizontal asymptotes happen if there is a limit to infinity and negative infinity.

Can you find limit as x goes to infinity and negative infinity?

Answer 2

I see only 1 horizontal asymptote of 1

Answer 3

That's odd.

Did you do
\[\lim_{x \rightarrow \infty} \frac{1-\frac{2}{x^2}}{2|1|+\frac{3}{x^3}} = \frac{1}{2}\]
\[\lim_{x \rightarrow -\infty} \frac{-x-\frac{2}{x}}{2|-x|+\frac{3}{x^2}} = \frac{-1}{2}\]

Answer 4

btw when x is negative. x^3 is also negative so that's why as x goes to negative infinity x^3 also becomes negative.

Answer 5

I see it on the graph and make only lists 1 lol but I see 1 and the 1/2 so it's 1/2 right? the answer 1/2

Answer 6

well actually the distance between 1/2 and -1/2 is 1

Answer 7

wait let me try to graph it

Answer 8

https://www.desmos.com/calculator/2rm55ga60r

Answer 9

see it now?

Answer 10

oh okay

Answer 11

alright. 2nd picture.

f(x) is continuous if there's no discontinuities.

Here the function splits apart when x = 3.

Basically when x=3. f(x) has to be the same for x+1 and x^2-2x + a

when x=3
x+1 = 4
Now you need to solve
4 = x^2-2x+a when x = 3

Answer 12

a is 1

Answer 13

ye

Answer 14

next one

Answer 15

okay this third one looks hard.

I know that a removable discontinuity happens when you have like
\[\frac{(x+a)(x-4)}{(x+a)}\] here there's a removable discontinuity at -a

hmmm

Answer 16

so -pi?

Answer 17

OH I got it.

If the denominator is 0 and the numerator is 0 then it's a removable discontinuity.

if x = -pi then it's sin(0)/0 so YES

Answer 18

if x = pi then it's also sin(0)/0 so pi and -pi both work

Answer 19

okay

Answer 20

but if x = 0. Then it'll be 0/-pi^2 so that is NOT a removable discontinuity. A

Answer 21

thank you so much! I shouldn't have too many more I hope and after today I'll be done with this old curriculum and may not need much help in the course!

Answer 22

okay so for the first one. You only need to know if it's positive or negative.
if f'(x) is positive. it's going up. if it's negative. It's going down.
f(x) just tells you where it's at.

if f(x) is positive and f'(x) is positive. It's like this:

if f(x) is positive and f'(x) is negative it's like this:

If f(x) is negative and f'(x) is positive It's like this:

if f(x) is negative and f'(x) is negative it's like this:


now just connect the dots to figure out when it crosses the x-axis

Answer 23

right?

Answer 24

Then if we connect them it looks like this.


it HAS to cross the x-axis 2 times.

Answer 25

so the question is basically asking how many times it crosses the x-axis?

Answer 26

yeah. cause when it crosses the x-axis. f(x)=0

Answer 27

so the answer is 2

Answer 28

ye

Answer 29

now f'

Answer 30

That's when the line is perfectly straight. It does that 2 times.

ignore the one on the very top that doesn't have to be there

Answer 31

so both are 2?

Answer 32

yeah

Answer 33

This 3rd one looks really hard. I don't know if I can help you on it

Answer 34

that's okay I'll figure it out

Answer 35

okay well it's basically
\[\frac{f(x+h)-f(x)}{h}\] but simplify all that crap

Answer 36

and I think the 4th picture is the same formula but this time h=0.5 and x = 2

Answer 37

That's just
\[(\sin(x))^{-1}\] use the chain rule where u = sin(x)
\[(u^{-1})'*(u)' = -u^{-2}*(\cos(x)) = -(\sin(x))^{-2}*\cos(x) = \frac{\cos(x)}{-\sin(x)^2}\]

Answer 38

and second derivative is derivative of that right?

Answer 39

yeah

Answer 40

For the first one:

Answer 41

if there is no f'(x) = 0 then maybe it's like this. As long as f'(x) is positive then it's at the very right of the interval. f(2).

Answer 42

is the max 6?

Answer 43

uh hold on
\[0 = 12x^2+12x-8, x=-1.457427107756338\]
f(-1.457427107756338) = 10.021 ish

Answer 44

They gotta give you a calculator for THAT problem. dang

Answer 45

virtual school lol

Answer 46

for the next problem. Find out where the graph changes direction.

Answer 47

So it should be like this

Answer 48

cool!

Answer 49

3rd picture is easy.

You gotta figure out how to connect the derivative of both functions
2x + 2 = 2ax for x = 1;

2(1) + 2 = 2a(1)
4 = 2a

a = 2

Answer 50

so d

Answer 51

ye. let me double check that with a graph

Answer 52

tada https://www.desmos.com/calculator/16bqh5sdo2

Answer 53

Okay last one:

if something is perpendicular, then the slope is inverted.

f'(2) = 2. So slope is 2. Invert that and slope is -1/2

now you know f(2) = 5 so

5 = (-1/2)x + b

find b when x = 2

Answer 54

so 4 so c ?

Answer 55

Actually b

Answer 56

5 = -(1/2) (2) + 6

Answer 57

oh right it's -1 so yeah

Answer 58

4 more questions? last 4

Answer 59

I'm tired bro. I gotta start studying me own stuff for my midterm exam tomorrow

Answer 60

okat ty for helping!