sketching graph + 2 functions questions

[TrueTears]

Q1. How would you sketch ? Well, the question is actually find the range of when ? I was thinking about sketching the graph and work it out from there, but i have never met something like . Is there a way to work out the range without sketching? If so, how would you sketch graphs like where in general ( without the use of a calc)?

Q2. Let a be a positive number, let and let . Find all values of a for which and both exist.

I've gotten this far: ie, for to exist. And , ie for to exist.

Also and .





How do you solve it from here? Or do you need to do it a completely different way?

Q3. Let . If b is the smallest real value such that g has an inverse function, find b and

Many thanks!

[bucket]

I just sketch them, but I did methods cas.

[dekoyl]

Did you do GMA?
Sketch
The x-intercepts indicate an asymptote as
So it looks like half a truncus (the right half of )

[TrueTears]

Did you do GMA?
Sketch
The x-intercepts indicate an asymptote as
So it looks like half a truncus (the right half of )

yeah i did GMA but we never learnt how to sketch 1/ sqrt (x) . i know how to sketch Then how do u work out the asymptotes for 1\sqrt (x)

[dekoyl]

yeah i did GMA but we never learnt how to sketch 1/ sqrt (x) . i know how to sketch Then how do u work out the asymptotes for 1\sqrt (x)

It involves logic as well.
When , that is the x-intercept, right?
Now, at the x intercept, knowing , the graph of would be . Because never touches x = 0 (because you know that it'll be at x =0), it will be an asymptote there.

Try sketching on your calculator. It might help you understand what I'm trying to say more:
y1 =
y2 =

Oh shit I suck at explaining Hope you get it.

[dekoyl]

Also not really required but..

at
because

and


So if you were to draw and on the same axis, they'll have a common point at

[TrueTears]

hmm yeah i kinda get that, like when i work out how to sketch hyperbolas or truncas what i do is: if i get something like
the asymptotes will be at x=2 and y=2 since the graph is shifted across 2 to the right and 2 up, then because i also know the general shape of a hyperbola, i'd find the x and y intercepts and sketch it. And if it is just . The asymptotes are just the y axis and x axis ie, x=0 y=0. So if you get what is the general shape of this family of graphs? Is it just the same as the 1st quadrant of a hyperbola except it gets steeper much faster?

So if you get \frac{1}{\sqrt{x+4}} -5. How would you work out the asymptotes here? Is it just x=-4 and y=-5. Then work out the y intercepts which = -4.5 and x intercept = -3.96 and sketch it with the general shape of a hyperbola except it gets steeper much faster?

Something like this?

[dekoyl]

So if you get \frac{1}{\sqrt{x+4}} -5. How would you work out the asymptotes here? Is it just x=-4 and y=-5. Then work out the y intercepts which = -4.5 and x intercept = -3.96 and sketch it with the general shape of a hyperbola except it gets steeper much faster?

Something like this?

(Image removed from quote.)

Yep that's right.
(But really, I don't think I've ever sketched it in methods.. or very rarely. I did this mostly in GMA)

[kurrymuncher]

Just think of it as a reciprocal. 1/sqr(x) is the same as sqr(X) flipped upside down.

[/0]

Try graphing graphs of for different values of . They all have the basic 'hyperbola' shape.

As gets bigger, the graph becomes steeper for and flatter for . The trend is kept for fractional powers, although if the index has an even number in the denominator, x cannot be negative, and if it has an even number in the numerator, y cannot be negative.



Q2

,  

So 1. we need to find so that any value of will only be a subset of

The function is a straight line and always decreasing, so the maximum value will be at its left endpoint, at . Hence, we can say . For it to have the correct range, we require

and 2. we need to find so that any value of will only be a subset of

For , determines the vertical translation of the standard parabola with vertex at (0,0). We can say . So to have the right range, .

Hence the answer is .


Q3

For a function to have an inverse, it must be one-to-one. For a quadratic function, the graph is one-to-one up till the vertex, and after that it becomes many-to-one.

For , the vertex is at , hence the smallest real value of is .

For the inverse,





But as , the sign has to be positive, so the inverse is

[TrueTears]

ah yes thank you all

And one more Q

A cuboid tank is open at the top and the internal dimensions of its base are x and 2x m. The height is h m. The volume of the tank ix V cubic metres and the volume is fixed. Let denote the internal surface area of the tank.

a) i. Find S in terms of x and h.
ii. Find S in terms of V and x.   so

This is all fine, just the next few i can't get

b) State the maximal domain for the function defined by the rule in a) ii.how do you find the implied domain here?
c) if 2<x<15 find the maximum value of S if V= 1000

[/0]

b)



Hmmm the only real restriction I can see here is that (from a practical aspect lengths must be positive).
x cannot equal 0 but that case is already covered

c)

To find the maximum of a graph in a given domain you need to do 2 things:
1. Check the endpoints.
2. Check stationary points for local maximums.
If possible, look at the graph.

,  

Differentiating with respect to x and solving equal to 0, stat. point is at , giving

So the maximum value of S is S = 1508. (although x cannot equal 2, the value x gives near 2 is so close to 1508 that the difference is negligible)

[TrueTears]

ah yes i was wondering why x = 2, because it said 2<x<15 so 2 is technically not included but yeah thanks

just another 3 questions

A1


if find . The answer is



where does the and come from?

Q2
Let . What is the range of f(x) ?

Q3
Let where . If for, find the rule for in terms of q

[/0]

Q1

We require

i.e. for the first part and for the second part.

Now you have to find the domain from this range, (its a bit like finding the range in reverse)

For , for the range , the domain must be restricted to
and for the range , the domain must be restricted to . And that's how you get the domains for the hybrid function.

Q2.

If you perform long division on the fraction, you get



Since are all constants, this is just a hyperbola. The asymptotes are at and .

So the range is .

Q3.

I don't really know how to do this question.... this is as far as I got.







So and

[TrueTears]

So, how would u define the domains here?



if find .

The is just R right? so the domains of f(x) won't change when u plug it in the hybrid equation of right?