Question

Extrema, finding x and y intercepts, points of inflection, etc. Please check my work so far and help with the rest! :)

Answer 1

@nuttyliaczar :) you free?

Answer 2

Question 8.

Answer 3

Could you possibly give a better quality picture of your work? I don't want to incorrectly check anything

Answer 4

sure, I am rewriting it now :) give me a few minutes.

Answer 5

No no the handwriting is great, but the camera might have been too far

Answer 6

Ah ok.Lemme see.

Answer 7

Sorry, I think this last one is sideways.

Answer 8

That's up to part c. I am still writing out part d but I'm almost done :)

Answer 9

You are doing a great job, but I think the simplification after your rationalization in part c had some mistakes

Answer 10

The right side is fine I think, but the (6-x) side should be looked at again

Answer 11

hmm so after I get the common denominator?
after the line with all the big parenthesis?

Answer 12

Yes the line that has 6-x as the numerator

Answer 13

I think you may have forgotten to distribute

Answer 14

It looks to me like you combined the two x's and made one full x, but you forgot the other terms there

Answer 15

is the result correct though? o.o
I'm pretty sure the result is correct so i may have just missed a number.

Answer 16

http://www.wolframalpha.com/input/?i=%286-x%5E%282%2F3%29%29*%286-x%29%5E%281%2F3%29

Answer 17

So unfortunately I do not think it is correct. But up till that line you were fine

Answer 18

hm ok. So what is the result? xP
he did that part in class so I didn't bother to check it but he may have done it incorrectly haha

Answer 19

Personally I wouldn't combine those two quantities. I would leave them in their parentheses

Answer 20

Okay. and just put it with parenthesis on the line that has 2x on the right side?

Answer 21

Yes

Answer 22

because in the next line I want the simplest form of f'(x)

Answer 23

To be safe I'll check the whole function on wolfram

Answer 24

ok :)

Answer 25

Hmm the solution for the derivative on wolfram is different but I'm not sure why. I was following your work and saw only that mistake

Answer 26

How i wrote it can be farther simplified.

Answer 27

Oh nvm yeah I see it now. Thanks for pointing that out

Answer 28

I think you had it right, I'll look at my own work again

Answer 29

Ohhhhh you dropped the parentheses earlier!

Answer 30

It makes sense again, I thought the ^2/3 was applying only to x but that's because you forgot parentheses

Answer 31

ooh shooot. It should be (6-x)^2/3 right?

Answer 32

Yeah

Answer 33

i'm so sorry. my bad xD

Answer 34

so once I put those in its all good?

Answer 35

I should have caught that too

Answer 36

Yeah then you're good. Make sure you simplify the last term by dividing out the 3

Answer 37

and if I do that I get
[2-x]/[x^(2/3)*(6-x)^1/3)] yeah?

Answer 38

But be careful with your roots. The bottom does not have to be 0

Answer 39

And yes that's correct

Answer 40

For the critical points I get
2 from the first one and 0,6 from the second

Answer 41

I don't think you need to find when the denominator is 0

Answer 42

When the denominator is 0, your function will skyrocket to infinity

Answer 43

So on that note I only see 2 as a critical point

Answer 44

brb i need to drive home xD
hmm ok. Without the 6 I get no minimum? have you looked at the graph?

Answer 45

I mean if you count -infinity as a minimum (which I never did) then yes I suppose that is

Answer 46

Those are what we call asymptotes. That's what the 0 represents too

Answer 47

haha idk man. the graph supports it as a minimum..

Answer 48

It works as a minimum because the vertical line is on the derivative, not on the original function. On the original function it is defined, the slope is undefined

Answer 49

Yeah yeah I forgot to go back to the original my bad

Answer 50

ok xD So ...all good?
I finided d and e now, shall I attach those?

Answer 51

Yes

Answer 52

ugh sideways, sorry :/

Answer 53

[ and the second one goes first]

Answer 54

I like both ways you did it. The first by using a limit and the second I'm guessing you used the fact that it's a polynomial?

Answer 55

Yeah, i should write "considering this function is a polynomial...." ?

Answer 56

which way is better? xP because they both say the same thing, no?

Answer 57

Keep both to impress your teacher ^^ and your inflection point work is good too

Answer 58

Really? you think? :o

Answer 59

yay!! it's all done then I think :}

Answer 60

Good job, it's been too long since I did this stuff so I apologize for the crude checking

Answer 61

thank you so much. haha no worries, it was good to double check my work and do second guessing xD
the dropped parenthesis would for sure be points off :)