Questions thread

[Hielly]

how to find the domain of this?

Find the inverse of each of the following functions, stating the domain and range of
each:
g: [1,∞) → R, g(x) = square root(x − 1)

so the inverse is y=x^2 +1

when i sketched this, its a parabola so i was thinking the domain is R

[qshyrn]

thanks

x=y^2 + 2y

make y the subject. How woule i do this?


thanks

y^2+2y-x=0, then use the quadratic formula to solve for y. (a=1 , b=2, c=-x)   ofcourse this is the same as TT's way without much working.

[qshyrn]

how to find the domain of this?

Find the inverse of each of the following functions, stating the domain and range of
each:
g: [1,∞) → R, g(x) = square root(x − 1)

so the inverse is y=x^2 +1

when i sketched this, its a parabola so i was thinking the domain is R

the domain of the inverse function is the range of the original one. the range of the original one was [0 , infitity) so therefore thats the domain of the inverse.

[Ilovemathsmeth]

Domain of the inverse is range of the original. Range here for the original is [0, infinity). Thus this is the domain for the inverse function. I don't think you need to go any further than that.

[Hielly]

squareroot(x)+2=y

To find the x-intercept i set y=0.
So,
squareroot(x)+2=0
squareroot(x)=-2
x=4

However when i graph this on my calc, there is no x intercept at 4..

Am i missing something?

Thanks!

[qshyrn]

squareroot(x)+2=y

To find the x-intercept i set y=0.
So,
squareroot(x)+2=0
squareroot(x)=-2
x=4

However when i graph this on my calc, there is no x intercept at 4..

Am i missing something?

Thanks!

there aint any x intercept.. (think of the graph y =sqroot x, then translate that up by 2 and youll see )
sqr root of 4 is not -2 , but only 2 (your mistake was that from squareroot(x)=-2, you assumed that x was 4)

[TrueTears]

When you square you are creating redundant solutions. A good way to check is sub your final answer back in and see if it makes sense.

[kenhung123]

A rectangular piece of cardboard has dimensions 20cm by 36cm. Four squares each x cm by x cm are cut from the corners. An open box is formed by folding up the flaps. Find a function for V, which gives the volume of the box in terms of x, and state the domain for the function.

[QuantumJG]

A rectangular piece of cardboard has dimensions 20cm by 36cm. Four squares each x cm by x cm are cut from the corners. An open box is formed by folding up the flaps. Find a function for V, which gives the volume of the box in terms of x, and state the domain for the function.

This is a very simple question! (Note: These are your gift questions in the methods exam 2)

First lets look at our cardboard,

its dimensions are 20cm x 36cm

we then cut four squares off each corner implying that our new dimensions are,

20-2x by 36-2x

Now imagine folding each of these parts up, you will get an rectangular box of

height (h) = x
length (l) = 36 - 2x
width (w) = 20 - 2x

Now we define the volume of the box (V) as,

V = h*l*w = x*(36-2x)*(20-2x) = x*2(18-x)*2(10-x) = 4x(18-x)(10-x)

Now get your graphics calculator (or even just sketch the graph) and look at the graph.

When x<0, V is negative and when 10<x<18, V is also negative.

Now you could potentially say that the domain can either be 0<x<10 or x>18 since we KNOW that you can't have negative volume, but, we also know that x can't be greater than 10cm since we would have no width to our box, it would just be a fold of paper.

So we can safely say that the domain is 0<x<10 or (both are fine)

that answers your question,

we also can find the range where,



You may also be asked for the dimensions where the Volume is at its maximum, this is easy!

Just find the value of x when , by the graphics calculator you will find that the value is x = 4.13cm,

So the height = 4.13cm
          length = 36 - 2*4.13 = 27.74cm
          width = 20 - 2*4.13 = 11.74cm

[kenhung123]

I did it on calculator but it came up with 4x(x-10)(x-18)

[QuantumJG]

V = 4x(18-x)(10-x) = 4x*(-1)(x-18)*(-1)*(x-10) = 4x*(-1)² *(x-18)(x-10) = 4x(x-18)(x-10)

[Ilovemathsmeth]

Often the negative sign is removed to produce a slightly different looking answer which is essentially the same.

[Hielly]

Find
{x:x^(3/2)>x^2}

So we sketch these two equations on the same graph, and look at the x and y-intercepts? but not sure what to look for?

[kenhung123]

Often the negative sign is removed to produce a slightly different looking answer which is essentially the same.

Won't that mean the final answer would be the opposite sign? i.e. +>- and ->+?

[QuantumJG]

Find
{x:x^(3/2)>x^2}

So we sketch these two equations on the same graph, and look at the x and y-intercepts? but not sure what to look for?

simple 0 < x < 1

EDIT: I will explain this a bit better,

Graph both functions,

set your axis: x-min = 0 x-max = 1.5
                   y-min = 0 y-max = 2

so if you look at the two graphs (y₁ = x^1.5, y₂ = x²), y1 is above y2 until x = 1 (at x=1, y₁ = y₂ )

If you want to solve this by hand, you can say that,

x^1.5 > x² x³ > x⁴ (just by squaring both sides) x³ - x⁴ > 0

which can be written as,

x³(1 - x) > 0, then you can say that this is true only when x<1, but greater than 0