Mathematics Question Thread

[fun_jirachi]

Hi! Sorry, I'll be more specific.

I found the stat pt and tried to find slope around it, except I got that the gradient is the same ? so is it an inflexion point?
as for limits, I got as x--> infinity, y--> - infinity and as x--> - infinity, y--> infinity. is there a way to figure these out without subbing in very large and very small numbers?

Out of curiosity, which points did you test? Because it's highly unlikely they were the same. But yes, it is an inflexion point as you can find by finding the second derivative (which I haven't bothered to add in, seems unnecessary.) Your limits are a bit off, as x tends to infinity, y should tend to zero (remember to sub into the original function, not the derivative). Subbing in big numbers is the easiest way, and I guess that's all you need to use in 2U. It's the easiest and quickest way to do so.

Sketch y=xlnx.

so i differentiated and got (1/e, -1/e) as a min TP, and no inflexion points; x--> infinity and y--> infinity, and x can't approach negative infinity because x can't be equal to 0 therefore there's an asymptote at x=0, is that right? I have looked at the answers and i just don't get the open circle thing at (0,0). There's an asymptote at x=0 and also an x-intercept at 0, why

There isn't an asymptote at x=0; there is an open endpoint. Remember that lnx is defined only for x>0. Because of this, there is an open circle at (0,0) because lnx isn't defined there. You assess the limiting behaviour of xlnx as x->0 and it because it approaches zero, you just let the 'endpoint' be (0,0). Also, xlnx only has negative values for 0<x<1 because over this domain y=x is positive, but y=lnx is negative --> you will never approach negative infinity. For all other x, lnx is positive as is x over the same domain, and thus it approaches infinity. Important to note there is not technically an asymptote or a x-intercept at x=0. (explained above due to graph being undefined there - an asymptote is a line of discontinuity: think y=1/x). Also note that having an x-intercept and a asymptote at the same point is impossible, they are two mutually exclusive things for any function. In fact, you can't have any point with an x-value corresponding to a vertical asymptote.

Hope this helps

[david.wang28]

Hello,
I am stuck on a challenging 2 unit question in the attachment below (I have no idea how to do it). Can anyone please help me out? Thanks

[jamonwindeyer]

Hello,
I am stuck on a challenging 2 unit question in the attachment below (I have no idea how to do it). Can anyone please help me out? Thanks

This looks impossible because there is an \(a\) in the answer but not in the given terms!!

[julz_roha]

Hello Atar Notes.
How do we use the first derivative to check for points of inflexion?
Please Reply.
 - Juliana.

[Bri MT]

Edit: I misinterpreted something due to differences across state's curriculms. Just go read Rui's response.

my original response

Hey Juliana,

To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)

2. Now we need to figure out what type of stationary point it is.  Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right. 

Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0  or f'(x)<0 on both sides of f'(x)

I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask

The above info means that we can test the nature of a stationary point by finding f'(x)  a little to the left & a little to the right of the x value you found in part 1.

Let me know if this is unclear

(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)

[RuiAce]

Hello Atar Notes.
How do we use the first derivative to check for points of inflexion?
Please Reply.
 - Juliana.

Hey Juliana,

To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)

2. Now we need to figure out what type of stationary point it is.  Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right. 

Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0  or f'(x)<0 on both sides of f'(x)

I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask

The above info means that we can test the nature of a stationary point by finding f'(x)  a little to the left & a little to the right of the x value you found in part 1.

Let me know if this is unclear

(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)

Hold. I would like to interfere here.

A point of inflexion is not at all related to the first derivative. The question provided does not flow logically because it contradicts basic principles.

Points of inflexion do not refer to points of horizontal tangents in general (although occasionally they can, see below). They refer to points where the concavity of the curve changes. By itself, the first derivative is useless for this. We require the second derivative to deduce possible points of inflexion.

In general, setting the second derivative equal to 0 gathers all such points for ya. THEN, we apply a change in concavity test (sub a point close to both sides of the possible inflexion point) to determine if it actually is a point of inflexion.

The first derivative only comes into play for the horizontal point of inflexion. This is a special case where both the first and second derivatives are zero. But again, the second derivative is necessary.

[julz_roha]

Hey Juliana,

To find a point of inflection for a function f(x):
1. Solve f'(x) = 0 for x
This gives you the stationary points (point of inflection, local maximums, and/or local minimums)

2. Now we need to figure out what type of stationary point it is.  Around a local minimum, the graph goes down then up. So we have: f'(x)<0 on the left, f'(x) = 0 at the centre, f'(x)>0 on the right. 

Around a point of inflection we have f'(x)>0 on both sides of f'(x) = 0  or f'(x)<0 on both sides of f'(x)

I'll leave it to you to see what the "rule" is for a local max but if you don't get it feel free to ask

The above info means that we can test the nature of a stationary point by finding f'(x)  a little to the left & a little to the right of the x value you found in part 1.

Let me know if this is unclear

(Note: I did VCE not HSC so I'm unsure if this method is part of your curriculm)

Hold. I would like to interfere here.

A point of inflexion is not at all related to the first derivative. The question provided does not flow logically because it contradicts basic principles.

Points of inflexion do not refer to points of horizontal tangents in general (although occasionally they can, see below). They refer to points where the concavity of the curve changes. By itself, the first derivative is useless for this. We require the second derivative to deduce possible points of inflexion.

In general, setting the second derivative equal to 0 gathers all such points for ya. THEN, we apply a change in concavity test (sub a point close to both sides of the possible inflexion point) to determine if it actually is a point of inflexion.

The first derivative only comes into play for the horizontal point of inflexion. This is a special case where both the first and second derivatives are zero. But again, the second derivative is necessary.

Thank you \ ^ o ^ /
I understand now.
 - Juliana.

[shaynec19]

Could someone please help me here.

A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
a) How much is each instalment?
b) How much is paid back altogether?

[david.wang28]

Could someone please help me here.

A loan of $6000 over 5 years at 15% p.a. interest, charged monthly, is paid back in 5 annual instalments.
a) How much is each instalment?
b) How much is paid back altogether?

For a), you want to start off with A1 = 6000[1+(0.15/12)] - M and then write a few similar lines to A1 until A60 (60 months in 5 years). Let A60 = 0, then do some simple rearranging. Then, you want to use the sum of GP formula, so you get M[(1.0125^60 - 1)/0.0125] = 6000(1.0125)^60. Then find M, then times 12 to get the answer.
For b), times the answer of a) by 5, and you should be there.
Hope this helps

[shaynec19]

For a), you want to start off with A1 = 6000[1+(0.15/12)] - M and then write a few similar lines to A1 until A60 (60 months in 5 years). Let A60 = 0, then do some simple rearranging. Then, you want to use the sum of GP formula, so you get M[(1.0125^60 - 1)/0.0125] = 6000(1.0125)^60. Then find M, then times 12 to get the answer.
For b), times the answer of a) by 5, and you should be there.
Hope this helps

That's what i did, and the answer is apparently wrong. It has got to do with the interest being charged monthly, but each installment being paid annually. From what I can work out, the r value of the series is 1.0125^12, but I can't seem to get the right answer.
Part a answer is $1835.68

[david.wang28]

That's what i did, and the answer is apparently wrong. It has got to do with the interest being charged monthly, but each installment being paid annually. From what I can work out, the r value of the series is 1.0125^12, but I can't seem to get the right answer.
Part a answer is $1835.68

If the power is 12, then you should be writing 1.15^12. Try using that and let me know if you have any problems

[spnmox]

Find the area bounded by the graph of f(x) = x(x-2)(x+1) and the x-axis

So I'm not sure if this is right: area=37/12, and if I were to evaluate the integral, it would be 5/12 - 8/3 = -9/4??

[david.wang28]

Find the area bounded by the graph of f(x) = x(x-2)(x+1) and the x-axis

So I'm not sure if this is right: area=37/12, and if I were to evaluate the integral, it would be 5/12 - 8/3 = -9/4??

You got the final answer right. But you evaluated the integral erroneously. the -8/3 part should be positive, by putting that as an absolute value. Drawing the graph out will help you out. Hope this helps

[spnmox]

You got the final answer right. But you evaluated the integral erroneously. the -8/3 part should be positive, by putting that as an absolute value. Drawing the graph out will help you out. Hope this helps

Wait but isn't evaluating the integral and finding the area different? The integral can be negative but the area has to have the absolute value signs? If I were to put abs value signs around the 8/3, then the integral would be the same as the area ?

[david.wang28]

Wait but isn't evaluating the integral and finding the area different? The integral can be negative but the area has to have the absolute value signs? If I were to put abs value signs around the 8/3, then the integral would be the same as the area ?

Yes, evaluating the integral is finding a certain region of the area using the limits given to you. The reason why you should put absolute value signs is for the area below the x-axis. Then the integral you find will be the correct area.