HSC Tips - Graphs (1 Viewer)

[suiyi_z]

mapping

hey when ur mapping something like y=In(sinx) what is the strategy?? i keep getting the limits and stuff wrong!
OR any mapping hints in genearl?

 

[McLake]

Originally posted by suiyi_z
hey when ur mapping something like y=In(sinx) what is the strategy?? i keep getting the limits and stuff wrong!
OR any mapping hints in genearl?

Check critical points (x=0, x->+/-infinity, any intercepts).

The use your calc to predict any values in regions your not to sure about and draw smootyh curves ...

 

[Affinity]

and other 'obvious' things such as periodicty and bounds

 

[aud]

Another tip: for those really strange graphs, like an ellipse on the diagonal in the middle of nowhere - implicitly derive the function, then let the denominator = 0, sub back into the original equation to find those points (which will now have vertical tangents); let the numerator = 0, repeat (will give points with horizontal tangents)... then just draw nice lines and axes everywhere and it should come out all pretty-like

 

[jarro_2783]

isn't finding where f''(x) = 0 sufficient to finding an inflection point? Because if you have found the turning points and checked their concavity with f''(x), then you know where f''(x) = 0 in between those two points will be an inflection point.

 

[McLake]

isn't finding where f''(x) = 0 sufficient to finding an inflection point? Because if you have found the turning points and checked their concavity with f''(x), then you know where f''(x) = 0 in between those two points will be an inflection point.

Let us be clear here:

(1) f'(x) = 0 means that we have a STATIONARY POINT (note that I have not said turning point)

(2a) f"(x) > 0 + (1) means concave UP turning point
(2b) f"(x) < 0 + (1) means concave DOWN turning point
(2c) f"(x) = 0 + (1) can mean UP TP, DOWN TP or inflextion, depending on the circumstance, so checking of further points for gradient (or using the thrd dervitive method) is VITAL

 

[m_isk]

Let us be clear here:

(1) f'(x) = 0 means that we have a STATIONARY POINT (note that I have not said turning point)

(2a) f"(x) > 0 + (1) means concave UP turning point
(2b) f"(x) < 0 + (1) means concave DOWN turning point
(2c) f"(x) = 0 + (1) can mean UP TP, DOWN TP or inflextion, depending on the circumstance, so checking of further points for gradient (or using the thrd dervitive method) is VITAL

third derivative??

 

[Slidey]

See this thread for an explanation:

http://www.boredofstudies.org/community/showthread.php?t=62306&page=1&pp=40

 

[Will Hunting]

Q: How about addition/division/multiplication/etc of 2 graphs? do we need to find the Points of inflexion, Maxima and Minima?

A: Yes (sorry, this changes some of the points I have made so far, but I think it is clear where it is appropriate to use calculas)

I disagree. Sketching both graphs, critical points on the resultant graph are easily found through inspection. Calculus brings inefficiency to the curve sketching process, and should be eschewed at all costs. Stick by your former argument, man.

 

[Riviet]

A useful trick and a very powerful one for finding the domain/range of many functions: simply make x or y the subject!
e.g: y=x², we know the range is y>0 but making x the subject:

x=+sqrt[y]

Now, from the square root, we see that y has to be greater than or equal to zero for this particular relation to exist.

e.g2: y=1/(x²+1)
yx²+y=1
x²=(1-y)/y
x=+sqrt[(1-y)/y]
So 1-y>0, provided y=/=0
.'. y<1, y=/=0.

Now go and try it out on functions!

 

[housemouse]

A useful trick and a very powerful one for finding the domain/range of many functions: simply make x or y the subject!
e.g: y=x², we know the range is y>0 but making x the subject:

x=+sqrt[y]

Now, from the square root, we see that y has to be greater than or equal to zero for this particular relation to exist.

(b)e.g2: y=1/(x²+1)
x²+y=1(/b)
x²=1-y
x=+sqrt(1-y)
So 1-y>0
.'. y<1

Now go and try it out on functions!

Did you multiply both sides by x^2 + 1????????????
because it seems like something is wrong

 

[Riviet]

Yeah I did, forgot the y in the x², fixed now.

 

[PF]

Q: critical points:
1. vertical tangent- gradient is +- (infinity) at a vertical tangent
2. angular point- gradient changes on each side of an angular point

are they correct? in arnolds book, it is stated that there are 4 different types of critical points, what are they?

lol we use arnold book too .... does anybody else find this book realli annoying?
4 types: 1) gradient is different on both sides of critical point
2) the point is at the end of a finite domain
3) discontinuity (but the graph exists on either side of this point so)
4) when you differentiate the function to attempt to find gradient at that point, the denominator of that function gives you a zero eg differentiate x^ (1/3) = 0.33 x^-0.67 thus denom = 0 at 0 and (0,0) is critical point

 

[bored of sc]

Sorry if this has been asked:

How do you know if there will be diagonal asymptotes on the graph? Is when the higher power of x is on the numerator and you perform a long division to get the equation of this diagonal asymptote? Hmm...

Also, is there another way of finding diagonal asymptotes?

 

[cyl123]

In rational functions, to find oblique/diagonal asymptotes, you divide top and bottom by highest power in DENOMINATOR and cross out parts which goes towards zero to get diagonal asymptote.

eg f(x) = x^3/(x^2 +1)
divide top and bottom by x^2 gives f(x) = x/(1+1/x^2)
So clearly the asymptote is y=x

 

[bored of sc]

In rational functions, to find oblique/diagonal asymptotes, you divide top and bottom by highest power in DENOMINATOR and cross out parts which goes towards zero to get diagonal asymptote.

eg f(x) = x^3/(x^2 +1)
divide top and bottom by x^2 gives f(x) = x/(1+1/x^2)
So clearly the asymptote is y=x

Oh yeah. I learn this. Cheers.

 

[turntaker]

Graphs iz 2 hard