# Thread: First Year Mathematics A (Differentiation & Linear Algebra)

1. ## Re: MATH1131 help thread

Why is this true: Wouldn't it be 1 real zero?  Reply With Quote

2. ## Re: MATH1131 help thread Originally Posted by Flop21 Why is this true: Wouldn't it be 1 real zero?
Why do you think it's 1 real zero?

I think we need more info  Reply With Quote

3. ## Re: MATH1131 help thread Originally Posted by parad0xica Why do you think it's 1 real zero?

I think we need more info
If it's increasing for one part, then decreasing the other part, then there's 1 point where there's a zero since it's begun to decrease. Where's the other zero?  Reply With Quote

4. ## Re: MATH1131 help thread Originally Posted by Flop21 If it's increasing for one part, then decreasing the other part, then there's 1 point where there's a zero since it's begun to decrease. Where's the other zero?
What if f: (1,4) -> R such that f(x) = 1000 - (x-3)^2? (no zero)

We need more info, can you show what example was above this?  Reply With Quote

5. ## Re: MATH1131 help thread Originally Posted by parad0xica What if f: (1,4) -> R such that f(x) = 1000 - (x-3)^2? (no zero)

We need more info, can you show what example was above this?   Reply With Quote

6. ## Re: MATH1131 help thread

Example says:

If function is strictly increasing on interval (a,b), then f has exactly one real zero in [a,b]. Note: f(a) and f(b) have opposite signs.

If function is strictly decreasing on interval (c,d), then f has exactly one real zero in [c,d]. Note: f(c) and f(d) have opposite signs.

_____

Our function is strictly increasing on (1,3), so it has exactly one real zero in [1,3]. Note: f(1) and f(3) have opposite signs.

Our function is strictly decreasing on (3,4), so it has exactly one real zero in [3,4]. Note: f(3) and f(4) have opposite signs.

Hence, our function has exactly two real zeros in the interval [1,4].  Reply With Quote

7. ## Re: MATH1131 help thread Originally Posted by Flop21 If it's increasing for one part, then decreasing the other part, then there's 1 point where there's a zero since it's begun to decrease. Where's the other zero?
This is not clear...  Reply With Quote

8. ## Re: MATH1131 help thread

Aren't you still in high school ? Originally Posted by parad0xica Example says:

If function is strictly increasing on interval (a,b), then f has exactly one real zero in [a,b]. Note: f(a) and f(b) have opposite signs.

If function is strictly decreasing on interval (c,d), then f has exactly one real zero in [c,d]. Note: f(c) and f(d) have opposite signs.

_____

Our function is strictly increasing on (1,3), so it has exactly one real zero in [1,3]. Note: f(1) and f(3) have opposite signs.

Our function is strictly decreasing on (3,4), so it has exactly one real zero in [3,4]. Note: f(3) and f(4) have opposite signs.

Hence, our function has exactly two real zeros in the interval [1,4].  Reply With Quote

9. ## Re: MATH1131 help thread Originally Posted by Zen2613 Aren't you still in high school ?
Why can't I know this if I'm in high school?

The concepts used in this question can be understood by anyone. It's just that the mathematical language can be frightening and difficult to comprehend but when a diagram pops up, the tunnel will become clear Not sure I'm a primary school or high school student or something else.. up to you to deduce and decide :P  Reply With Quote

10. ## Re: MATH1131 help thread Originally Posted by Zen2613 Aren't you still in high school ?
Huh? I was getting a uni student vibe out of him for a while now.  Reply With Quote

11. ## Re: MATH1131 help thread Why are the first 2 answers not correct?

g'(x) = 3x-24x+45, so g(2) = 3 right?

and

h'(x) = ln(x-1), so h(2) = 0 ?

Or am I doing something wrong here?  Reply With Quote

12. ## Re: MATH1131 help thread Originally Posted by Flop21  Why are the first 2 answers not correct?

g'(x) = 3x-24x+45, so g(2) = 3 right?

and

h'(x) = ln(x-1), so h(2) = 0 ?

Or am I doing something wrong here?
That derivative g' looks peculiar...

Let's see the entire question  Reply With Quote

13. ## Re: MATH1131 help thread

Where is the mathsoc solutions to the calculus test 2 past papers??? I can find the solutions for every other test just not calc test 2??

Pls help.  Reply With Quote

14. ## Re: MATH1131 help thread Originally Posted by Flop21 Where is the mathsoc solutions to the calculus test 2 past papers??? I can find the solutions for every other test just not calc test 2??

Pls help.
Yeah me neither. Either the 3rd-5th years are too busy or just forgot about them.

Not sure who's gonna look after 1131/41 but I'm just gonna upload my 1151 solutions if I get them out...  Reply With Quote

15. ## Re: MATH1131 help thread Originally Posted by leehuan Yeah me neither. Either the 3rd-5th years are too busy or just forgot about them.

Not sure who's gonna look after 1131/41 but I'm just gonna upload my 1151 solutions if I get them out...
FML

The solutions mathsoc had for the other tests were so good for study haha, they full on explained each answer.  Reply With Quote

16. ## Re: MATH1131 help thread Originally Posted by Flop21 FML

The solutions mathsoc had for the other tests were so good for study haha, they full on explained each answer.
Ikr. But don't complain lol they didn't even have to do them

Just like InteGrand doesn't exactly have to help me but I'll always be grateful of it  Reply With Quote

17. ## Re: MATH1131 help thread

Just confirming I'm correctly using the definition of the derivative:

So to show that f(x) = x^2 then f'(x) = 2x, you get the formula, sub in and after expanding and simplifying you get 2x+h. Now as h -> 0, h = 0, so we make h = 0 and then we are left with 2x.

Is this how you use it? I.e. finding the limit as h -> 0, which is 0...? or???  Reply With Quote

18. ## Re: MATH1131 help thread Originally Posted by Flop21 Just confirming I'm correctly using the definition of the derivative:

So to show that f(x) = x^2 then f'(x) = 2x, you get the formula, sub in and after expanding and simplifying you get 2x+h. Now as h -> 0, h = 0, so we make h = 0 and then we are left with 2x.

Is this how you use it? I.e. finding the limit as h -> 0, which is 0...? or???
Correct. Basically if you had to blindly show that you substitute the formula into the definition of the derivative, and from memory 2x+h is a correct step.  Reply With Quote

19. ## Re: MATH1131 help thread Originally Posted by Flop21 Just confirming I'm correctly using the definition of the derivative:

So to show that f(x) = x^2 then f'(x) = 2x, you get the formula, sub in and after expanding and simplifying you get 2x+h. Now as h -> 0, h = 0, so we make h = 0 and then we are left with 2x.

Is this how you use it? I.e. finding the limit as h -> 0, which is 0...? or???
Yeah, that's how you get 2x. (This is something done in 2U I'm pretty sure.)  Reply With Quote

20. ## Re: MATH1131 help thread Originally Posted by InteGrand Yeah, that's how you get 2x. (This is something done in 2U I'm pretty sure.)
Really? I feel like it's kind of familiar... but don't remember ever actually being asked to use it.  Reply With Quote

21. ## Re: MATH1131 help thread Originally Posted by Flop21 Really? I feel like it's kind of familiar... but don't remember ever actually being asked to use it.
Finding derivatives from first principles is in the 2U syllabus, surely? And if it is, finding the derivative of x^2 would be a typical example.  Reply With Quote

22. ## Re: MATH1131 help thread Originally Posted by InteGrand Finding derivatives from first principles is in the 2U syllabus, surely? And if it is, finding the derivative of x^2 would be a typical example.
Hehe you're right. Just saw some examples in last year's 2u marathon.

You can tell there were certain things in 2u I just didn't pay attention to.  Reply With Quote

23. ## Re: MATH1131 help thread Originally Posted by Flop21 Hehe you're right. Just saw some examples in last year's 2u marathon.

You can tell there were certain things in 2u I just didn't pay attention to.
Tbh, almost all of my cohort just ignored it and counted on the fact it wouldn't be examined in the paper.  Reply With Quote

24. ## Re: MATH1131 help thread

f(x) = x|x|

If it exists, evaluate lim h->0+ [f(0+h) - f(0)] / h

What do I even do here? Why have they got 0 already in the formula?  Reply With Quote

25. ## Re: MATH1131 help thread

They want you to find lim h->0 f'(0)

Note that x=0 does not strictly imply the same thing as h=0
Also note that f(0) = 0

So if you want to find lim h->0+ f(h)/h

Note that as h approaches 0 from the right, x = x and |x| = x
(Instead if h approaches from the 0 left, x = x but |x| = -x)

So they want you to see if there's an answer to lim h->0+ h|h|/h

(Apologies for poor wording)  Reply With Quote

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