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C3 Modulus/Functions HELP



1) How did they find out that the asymptote for f(x) is -k?

2) How did they find out that the range is f(x)eR? And what would the range be if it was an decreasing function? (I don't understand the relationship between xeR and increasing functions)

(This is the question if you need to refer to it):

k.png31.4KB
Original post by creativebuzz


1) How did they find out that the asymptote for f(x) is -k?

2) How did they find out that the range is f(x)eR? And what would the range be if it was an decreasing function? (I don't understand the relationship between xeR and increasing functions)

(This is the question if you need to refer to it):


Are you asking how to draw it?
Original post by ThatGuyRik
Are you asking how to draw it?


Not exactly how to draw it but, as mentioned in my first post, how to find the asymptote of of f(x) = ln(x+k)
Original post by creativebuzz
...


Note ln(x)\text{ln}(x) is undefined when x=0x=0. That is, it has an asymptote at x=0x=0.

Likewise, consider ln(x+1)\text{ln}(x+1), when is this undefined? Well it's clearly when x=−1x=-1 because then we would have ln(0)\text{ln}(0). Hence we have an asymptote at x=−1x=-1.

Same argument for ln(x+k)\text{ln}(x+k), it is undefined when x=−kx=-k therefore we have an asymptote at x=−kx=-k

Also, you should see via a sketch that ln(x)\text{ln}(x) has a range over all the real numbers. But ln(x+k)\text{ln}(x+k) only translate the graph in some horizontal direction, and doesn't affect the range.
(edited 9 years ago)
Original post by creativebuzz
Not exactly how to draw it but, as mentioned in my first post, how to find the asymptote of of f(x) = ln(x+k)


Okay so we know where y=ln(x) crosses the coordinate axis at (1,0) if we were to translate it to ln(x+k) the you would be moving to the negative region ie to the left. Therefore the asymphtote is y=-k it also tells us x>-k So the line cannot go below the x value of -k.
^^ the guy two posts above mine answered your first question..

part b) the range is xER, because every single value of X that's involved in the graph; beyond the '-k' bit, have values. this is because it is between the regions of : 'x>-k'. that is, after '1-k, 0' on the graph, all the values of x all correspond to a single point on the curvy bit of the graph.

i'll explain (or try to):

Going by the guys' above logic (i rated it '+1' because he is right); then, as, -k, is the asymptote and a value on the x-axis, then every single value (of this particular graph) of x is mapped onto this particular graph.

consider the graph: y=x.

all the values of x; have one value on the line bit, right?

going back to this question, this graph has a value for each part of the curve bit. therefore the range becomes: for the region x>-k; all the values of x have a value on the curvy bit of the graph..

does that help? or have i just muddled your mind more? :/
(edited 9 years ago)
Original post by SamKeene
Note ln(x)\text{ln}(x) is undefined when x=0x=0. That is, it has an asymptote at x=0x=0.

Likewise, consider ln(x+1)\text{ln}(x+1), when is this undefined? Well it's clearly when x=−1x=-1 because then we would have ln(0)\text{ln}(0). Hence we have an asymptote at x=−1x=-1.

Same argument for ln(x+k)\text{ln}(x+k), it is undefined when x=−kx=-k therefore we have an asymptote at x=−kx=-k

Also, you should see via a sketch that ln(x)\text{ln}(x) has a range over all the real numbers. But ln(x+k)\text{ln}(x+k) only translate the graph in some horizontal direction, and doesn't affect the range.


Ah thanks, you explained that really well!

So what would the range be if it was an decreasing function? Would it still be f(x)eR because it still covers all real numbers..
Original post by creativebuzz
Ah thanks, you explained that really well!

So what would the range be if it was an decreasing function? Would it still be f(x)eR because it still covers all real numbers..



Consider f(x)=−xf(x)=-x which is a decreasing function, but that doesn't stop it's range from being the entire set of real numbers. When looking at the range, you have to consider any values that the function cannot take. It doesn't have anything to do with the function being increasing/decreasing.
Original post by creativebuzz
Ah thanks, you explained that really well!

So what would the range be if it was an decreasing function? Would it still be f(x)eR because it still covers all real numbers..


For ln(x+k)    x>−kln(x+k) \ \ \ \ x>-k this is never a decreasing function. Since the derivative of this at any point in the domain is positive:

1x+k\displaystyle \frac{1}{x+k} is always positive for x>−kx>-k, therefore ln(x+k)ln(x+k) is always increasing.

If you had −ln(x+k)    x>−k-ln(x+k) \ \ \ \ x>-k then it is always decreasing, but still the range is unchanged since −ln(x)-ln(x) has a range of all real numbers and then we follow the same reasoning in my last post. (Horizontal translation doesn't change the range)
Original post by SamKeene
For ln(x+k)    x>−kln(x+k) \ \ \ \ x>-k this is never a decreasing function. Since the derivative of this at any point in the domain is positive:

1x+k\displaystyle \frac{1}{x+k} is always positive for x>−kx>-k, therefore ln(x+k)ln(x+k) is always increasing.

If you had −ln(x+k)    x>−k-ln(x+k) \ \ \ \ x>-k then it is always decreasing, but still the range is unchanged since −ln(x)-ln(x) has a range of all real numbers and then we follow the same reasoning in my last post. (Horizontal translation doesn't change the range)


Ah that makes sense, thank you!

Would you mind explaining why they made u=-10 in part b when speed can't be negative and in part a I showed that u=10 not -10.



(this is the question if you need to refer to it):

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Original post by creativebuzz
Ah that makes sense, thank you!

Would you mind explaining why they made u=-10 in part b when speed can't be negative and in part a I showed that u=10 not -10.


They took the +ve direction as downwards so the initial speed u (when its going up) is negative, and the final speed (v) is positive.
Original post by SamKeene
They took the +ve direction as downwards so the initial speed u (when its going up) is negative, and the final speed (v) is positive.


but I thought that speed can never be negative... :confused::s-smilie:
Original post by creativebuzz
but I thought that speed can never be negative... :confused::s-smilie:


U and V are velocities, which can have direction.
Original post by SamKeene
U and V are velocities, which can have direction.


Ah that makes sense, thanks!

For this question, why is it (mu)(5g) instead of (2mu)(5g) because the question says that friction was applied to each particle, no?




This is the question if you need to refer to it:

Untitled.png16.4KB
Untitled.png54.6KB
Original post by creativebuzz
Ah that makes sense, thanks!

For this question, why is it (mu)(5g) instead of (2mu)(5g) because the question says that friction was applied to each particle, no?

Because they are treating both particles as one big particle, you can see this as they say they resolve for the entire system. The co-efficient of friction is still the same regardless of the mass of the particle.

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