Hard's methods questions (easy)

[hard]

Evaluating the followings limits

If f(x) = x-2/x+2 continuos at x=2? give reasons for answer

For f(x)=x3 [cubed] - 3x find {x:f'(x) = 0}

Can you please show me how you work these out because i'm a bit confused due to not having enough time to commit to methods due to 3&4's.

Any help would be appreciated. thanks in advance.

[Damo17]

Evaluating the followings limits

If f(x) = x-2/x+2 continuos at x=2? give reasons for answer

For f(x)=x3 [cubed] - 3x find {x:f'(x) = 0}

Can you please show me how you work these out because i'm a bit confused due to not having enough time to commit to methods due to 3&4's.

Any help would be appreciated. thanks in advance.

Not sure how to do the first one.

The second:


When

[vce08]

Two answers as you take the root of 1. Hence 1 and -1

[danieltennis]

Can you please show me how you work these out because i'm a bit confused due to not having enough time to commit to methods due to 3&4's.

Maths Methods 1+2 > Legal studies + VCD 3+4

[Damo17]

Two answers as you take the root of 1. Hence 1 and -1

True, i'm too tired.

[Mao]

for a function to be continuous at a, we have three conditions:
1. exists
2. exists
3.

here, , to verify continuity at



We evaluate by first trying to substitute the 2 in the limit, which exists (no nasty /0 stuff) and is 0, so

[enwiabe]

mao, you need 4 conditions.

You need the limit as x approaches a from the positive direction AND from the neg. direction

[dcc]

My frustration at being beaten by 2 minutes cannot be understated!

lol

[Mao]

mao, you need 4 conditions.

You need the limit as x approaches a from the positive direction AND from the neg. direction

actually, not quite.

the condition of a limit existing is that
1. the left hand limit exists
2. the right hand limit exists
3. the left hand limit = right hand limit

the condition of continuity at a point is that
1. the function exists
2. the limit exists at that point [thus encompassing all three conditions of the prior. By not specifying a direction of approach, it is assumed that the prior conditions are satisfied, that is left = right. but it is not within the scope of MM to show this ]
3. the function = the limit

PS: sorry dcc

[hard]

honestly you guys are legends! seriously. oh and to danieltennis, maybe in terms of importance next year but not this year.

oh yes and one more guys.

For f: R-->R where f(x) = x2 - 3x - 10 find:

a) the gradient where it crosses i) the y-axis ii) the x-axis


b)the coordinates of the point where i)the tangent is parallel to the x-axis ii)the tangent is parallel to the line y + 3x+4 = 0

once again i love you people
thanks heaps
i'll get straight to methods after 3&4 exams

[Mao]

it crosses the y axis when x=0:





it crosses the x axis when y=0:

substituting, the two tangents are


b)
when it is parallel to the x axis, it is horizontal, i.e. gradient = 0



hence the coordinate is

when it is parallel to , the tangent has a gradient of -3.

hence the coordinate is

[Collin Li]

For f: R-->R where f(x) = x2 - 3x - 10 find:

a) the gradient where it crosses i) the y-axis ii) the x-axis

b) the coordinates of the point where i)the tangent is parallel to the x-axis ii)the tangent is parallel to the line y + 3x+4 = 0

Hmm, yeah I thought it was weirded strangely until I saw Mao's solution.

"The gradient where it crosses the y-axis" means the gradient where crosses the y-axis, because by "it", they mean - the question should be clearer about that.

Don't need to find the tangents.

[hard]

thanks Mao!

sorry for not wording it correctly.

also can you check if i have done this right:



        

       
         

       
         

       
         

       
         

 
         

hence 
         
Note: this is using the first principle rules and we are expected to work out the derivative

[Mao]

mmmm not quite

Syntax on limits e.g. :

Code: [Select]

\lim_{h\to 0}f(x)
Syntax on fractions e.g.

Code: [Select]

\frac{f(x+h)-f(x)}{h}

[Damo17]

        

      
        

       
         

       
         

       
         

 
         

hence           

If you are not sure if you have got it right, look at and work it out using the short way.
Therefore