More Graphing Help Needed! (1 Viewer)

[blackops23]

Hi, I have about 6 or so question needed help with, so please bear with me...

Ok:

Q1: y=e^x - e^(-x)
(i) determine whether it's odd or even
IT'S ODD
(ii) determine it's gradient, find turning points.
How do I do this?
(iii) then sketch the graph

I sketched the graph, but the only way I did it was by drawing y=e^x and y=e^(-x), and subtracting the y coordinates from each other using my fingers --> is there a more efficient way to sketch it??
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Q2: y=1/(e^x + e^(-x))
(i) function is EVEN
(ii) Derivative is (e^(-x) - e^x)/[(e^x + e^-x)^2] --- got this from yahoo answers, still don't know how to derive it...
MAXIMA AT (0,0.5)

(iii) sketch the graph
Ok I managed to sketch it, here's what i did:
1. draw e^x and e^-x
2. add y-values together manually to get e^x + e^-x
3. then draw the reciprocal graph.
Is there an easier method that considerably cuts down steps 1 and 2??

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Q3: y=(e^x - e^-x)/(e^x + e^-x)
(i) ODD FUNCTION
(ii) As x--> infinity, y--> 1, As x --> negative infinity, y--> -1, therefore asymptotes at y=1, y=-1
(iii) Once again from google, derivative was found, there are no turning points,. but at the x-intercept (0,0), gradient is = 1, HOW DO I DERIVE THE DERIVATIVE?
(iv) Sketch the graph
asymptotes, gradient = 1 at (0,0) function is odd, made it pretty easy to sketch.

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Now for the harder ones...

Q4: (x^2)(y^2)=x^2 + y^2 --> Apparently it's a traffic officer at point duty... wtf


Q5: y^2 = x^4 - x^6 --- don't really know how to sketch y=x^4 - x^6, could use help with that.

I kinda missed out on implicit differentiation in class, so some help would be good. All I know is derive y as normal and multiply it with y'....


Thanks guys

 

[Bored Of Fail 2]

Q1 (ii) determine it's gradient, find turning points.
How do I do this?


y=e^x - e^(-x)

you do four unit maths and you cannot differentiate this?? :|

dy/dx = e^(x) +e^(-x) =0 for turning points

ie e^(x) + 1/ ( e^(x) ) =0 ( multiply throguh by e^(x) )

[e^(x) ]^2 +1 =0

etc

 

[blackops23]

yeah, i don't know = (

but how is d(e^x)/dy = e^x
and how is d(e^-x)/dy = -e^(-x) how does the derivation process come about?

 

[deterministic]

Have you dealt with differentiating exponentials in 3u yet???

d(e^x)/dx=e^x by definition.... if you dont know use your standard integral going right to left, which is differentiation.

 

[deterministic]

Now for the harder ones...

Q4: (x^2)(y^2)=x^2 + y^2 --> Apparently it's a traffic officer at point duty... wtf

Q5: y^2 = x^4 - x^6 --- don't really know how to sketch y=x^4 - x^6, could use help with that.

I kinda missed out on implicit differentiation in class, so some help would be good. All I know is derive y as normal and multiply it with y'....

Thanks guys

Implicit differentiation is really based on the chain rule learnt in 2 unit. consider y^2

d(y^2)/dx = d(y^2)/dy * dy/dx (Chain rule)
dy^2/dy= 2y so hence d(y^2)/dx=2y(dy/dx)

similarly for other functions of y.

Q4: rearrange so that:

(x^2)(y^2)-y^2=x^2
y^2(x^2-1)=x^2
y^2=x^2/(x^2-1)

sketch y=x^2/(x^2-1) first and then use the tips i told you in the other thread to get y^2

Q5:
y=x^4-x^6=(x^4)(1-x^2)=(x^4)(x-1)(x+1)
now you have all the zeros so it is just sketching a polynomial.

 

[blackops23]

Implicit differentiation is really based on the chain rule learnt in 2 unit. consider y^2

d(y^2)/dx = d(y^2)/dy * dy/dx (Chain rule)
dy^2/dy= 2y so hence d(y^2)/dx=2y(dy/dx)

similarly for other functions of y.

Q4: rearrange so that:

(x^2)(y^2)-y^2=x^2
y^2(x^2-1)=x^2
y^2=x^2/(x^2-1)

sketch y=x^2/(x^2-1) first and then use the tips i told you in the other thread to get y^2

Q5:
y=x^4-x^6=(x^4)(1-x^2)=(x^4)(x-1)(x+1)
now you have all the zeros so it is just sketching a polynomial.

Awesome.

About q1 and q2, is that the most quickest way to sketch the graphs?

 

[deterministic]

q1:
- obviously it passes through origin (odd function)
- after differentiating it turns out it has no turning points. so just notice as x-> inf, the e^-x becomes insignificant, so that the curve in the first quadrant should look like e^x. reflect that about origin (odd function)

q2:
- no vertical asymptotes
- as x-> inf, y->0 ; x-> -inf, y-> 0
-even function
- maximum at (0, 0.5), that really should given you an idea on what the graph looks like

 

[blackops23]

q1:
-as x-> inf, the e^-x becomes insignificant, so that the curve in the first quadrant should look like e^x.

Thanks deterministic, that tip really helped out, got q4 just a curve approaching y=1 and x=1, reflected in both x and y-axis, and q5 was a bowtie,

Here's another one that drove me bonkers, basically there were three parts
(1). Sketch y=2x/(x^2+1) --> odd function turning points (1,1) (-1,-1), curve approaches y=0 as x-> +inf and x-> -inf.

(2). Sketch the cubic function of (1), y= 8x^3/(x^2+1)^3 --> same as (1) almost same as one, horizontal inflexion point at (0,0) etc.

3. Sketch y=x^4/[(x^2+1)^4] without any further calculation, and with reference to part (1) and part (2)

Any help on this would be extremely appreciated

 

[deterministic]

(1) - denominator is always positive (square + positive no.) so no vertical asymptotes
- odd function
- y=0 -> x=0 => intercept (0,0)
- as x-> infinity, y-> 0, same for x-> -inf (horizontal asymptote of y=0)
- you can find turning points if you want, it helps when observing shape of graph.
Thus it should look like a very squashed s shape passing through from third quadrant (near the x axis), then curve away, curve back through the origin, then curve towards the x axis in the first quadrant.

(2)this is simply sketching the cube of (1). The idea behind four unit graphs is that given a graph of a function f(x), you should be able to sketch the graph of functions of f(x) such as f(x)^3, f(x)^2, 1/f(x), sqrt(f(x)) etc without refering back to the explicit equation of f(x) (you can always do that but its slower). It is important to note how the critical features- turning points, intercepts and asymptotes - are transformed. I advise you learn these, either through teachers or trying to derive them yourself. I shall give you and example:

Suppose I have f(x) already. I want to sketch y= f(x)^3. So I note:
- if f(x) is positive, cubing it will keep it positive. similarly if f(x) is negative. NOTE f(x) and f(x)^3 have the same shape (if f(x) increase, f(x)^3 increase and vice versa)
- INTERCEPTS: f(x)=0 implies f(x)^3=0, so the x intercepts of f(x) and f(x)^3 are the same
-TURNING POINTS: y'=d f(x)^3/dx = 3f(x)^2 * f'(x) (remember implicit differentiation)
This tells us that f(x)=0 implies y'=0, f'(x)=0 implies y'=0. So the x values of turning points of y=f(x)^3 are those of f(x) AND the intercepts of f(x) as well. The nature is easily determined by the shape of the graph.
-ASYMPTOTES: vertical ones are shared, horizontal ones are different:
eg. suppose as x-> inf, f(x)-> 5, then f(x)^3->5^3=125
This should all be enough to sketch the curve.

IF ALL ELSE FAILS, just plot some points maually if you have the equation of f(x) to begin with

 

[nelsonzheng]

1) y=2x/(x^2+1)

3) y=x^4/[(x^2+1)^4]

Your basically applying ^4 to graph 1

 

[blackops23]

umm could you please tell me how to actually sketch the graph say like in a exam, with reference to the two other graphs and without any further calculation, thanks

 

[NihaoMa]

you find the asymtotes and certain points such as x/y intercepts. then just draw it. I love maths.

 

[nelsonzheng]

umm could you please tell me how to actually sketch the graph say like in a exam, with reference to the two other graphs and without any further calculation, thanks

Given that you have already plotted graph 1, you can see the relationship between graph 1 and graph 3:

 

[blackops23]

what makes this graph so hard to me is that it isn't like the other questions in the question, i..e it isn't (GRAPH1)^4,

GRAPH1 ^4 WOULD have been 16x^4/(x^2+1)^4
this one doesn't have a 16 in it...

Also the "no further calculation bit" is killing me, how do I sketch it without using calculus???

 

[nelsonzheng]

what makes this graph so hard to me is that it isn't like the other questions in the question, i..e it isn't (GRAPH1)^4,

GRAPH1 ^4 WOULD have been 16x^4/(x^2+1)^4
this one doesn't have a 16 in it...

Also the "no further calculation bit" is killing me, how do I sketch it without using calculus???

Blue line: original plot
Red line: original plot divided by 2, thus half the amplitude
Yellow line: (Red line)^4
Thus, no 16.
You can find the maximums just by subbing in some numbers into the calculations u did for graph 1. then just guess the general shape of the curve

 

[blackops23]

Blue line: original plot
Red line: original plot divided by 2, thus half the amplitude
Yellow line: (Red line)^4
Thus, no 16.
You can find the maximums just by subbing in some numbers into the calculations u did for graph 1. then just guess the general shape of the curve

Thanks mate! The halving of graph 1 really helped me out!

But there's one more problem, it is much much simpler than any graph i've had a problem with today, but my method is probably wrong.
Ok heres the question:

Q. i) Discuss the beahviour of the curve at the neigbourhood of x=1 and for very large values of x.
ii) Find y', hence, coordinates of any turning points
iii) Sketch the curve

graph: y=x+1+[1/(1-x)]
THe problem i'm, having is discussing the behaviour of the curve near x=1, i.e. finding the asympote

Unlike the other questions in the question, this graph starts with "x+1"

The other graphs were like e.g. y=x-1+[(1/x-1)]. So like for this graph, as x--> 1, x-1 vanishes and y=1/(x-1) is the asymptote, i.e. x=1

However for this graph as x--> 1, NOTHING GETS ELIMINATED, all i get is y=2 + 1/(x-1)
You can easily see that as x--> + infinity, 1/x-1 goes away, so the asymptote is y=x-1

But as for the other asymptote, I can't make heads or tails how the graph acts as x-->1 from positive and negative sides.

Thanks guys

 

[jyu]

For y=x+1+[1/(1-x)], as x --> 1-, 1/(1-x) --> + inf, .: y --> + inf; as x --> 1+, 1/(1-x) --> - inf, .: y --> - inf.

Don't see your problem!

 

[blackops23]

no worries, sorry got things mixed up, graph was 1/1-x, I had 1/x-1 in my head. So I guess the asymptote is y=2 + 1/(1-x)

 

[nelsonzheng]

However for this graph as x--> 1, NOTHING GETS ELIMINATED, all i get is y=2 + 1/(x-1)

y=x+1+1/(x-1)

Let's split this graph into y=x+1 and y=1/(x-1)
y=x+1 is a straight diagonal line with y intercept of 1 and a positive gradient. (this doesn't really change the question)
y=1/(x-1) is a hyperbola with an asymptote at x=1
Then you add the 2 together as seen below:

Hope that helps.

 

[jyu]

Hi Nelson, you misread 1/(x-1) for 1/(1-x)