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Hi EP,

I appologise EP your answers are correct.  

You probably do not need to worry about this explaination, I expect you already know it.

It might hlp the guest that asked though.


What I was getting at is this  (but it makes no differnence for this question)


The definition of a function is that for every x vlaue there can be at most 1 y value.

If it is graphed this is easy to see from the straight line test.


The inverse of a funtion is its reflection in the line y=x

When you do this, most times the reflection is NOT a function, so you need to be very careful about the new domain.


All the teachers I have ever discussed this with say you must rearrange the equation to make x the subject BEFORE you swap the x with the y BECAUSE then the new domain and range are more easily seen.


I cannot see why they make this demand so strongy but certainly people must be extremely careful about constraints.


say if

y=x^2  this is a normal concave up parabola

If you just swap the lpronumerals like you said then you get   x=y^2    This is a sideways parabola BUT it is not the inverse function of y=x^2 because it is NOT a function at all.


To find the inverse of y=x^2 you should rearange the formula first and make sure that for each y value there is ONLY one x.


THEN you can swap the letters over.


\(y=x^2\\ \pm\sqrt{y}=x\\ x=\pm\sqrt{y}\\ \text{It can easily be seen that } y\ge0\\ \text{Swap pronumerals and drop the minus sign}\\ y=\sqrt x \qquad \text{If you include both answers this will not be a function }\\ \text{The inverse of }y=x^2 \text{ is } y=\sqrt{x} \quad x\ge 0 ,\quad y\ge0 \)



Here is the graph of y=x^2 in red and its inverse function in blue


Melody 22.02.2018

I am quite sure EP will be correct but I will take a look:


Let b and c be constants such that the quadratic \(-2x^2 +bx +c\)   

has roots \(3+\sqrt{5}\)  and \(3-\sqrt{5}\).

Find the vertex of the graph of the equation   

\(y=-2x^2 + bx + c\)


The roots of a quadratic   \( ax^2+bx+c\)  is the answers to   \( ax^2+bx+c=0\)

And the answers to this are given by the quadratic equation


\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=3\pm\sqrt5 \)


This means that  \(x=\frac{-b}{2a}\) = 3      must be exactly halfway between the two roots.!!


So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3

The y value will be    \(y=-2*3^2+bx+c = -18+3b+c \)


This is fine but you need to find the vleu of b and c       frown


From the equation above I can see that 

\(\frac{-b}{2a}=3 \qquad and \qquad \pm\frac{\sqrt{b^2-4ac}}{2a}=\pm\sqrt5\\ a=-2\qquad so\\ \frac{-b}{-4}=3 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{-4}=\pm\sqrt5\\ b=12 \qquad and \qquad \pm\frac{\sqrt{b^2+8c}}{4}=\pm\sqrt5\\ b=12 \qquad and \qquad \frac{b^2+8c}{16}=5\\ b=12 \qquad and \qquad b^2+8c=80\\ b=12 \qquad and \qquad 144+8c=80\\ b=12 \qquad and \qquad 18+c=10\\ b=12 \qquad and \qquad c=-8\\\)




\(y= -18+3b+c\\ y= -18+3*12+-8\\ y=-18+36-8\\ y=10 \)


So the vertex is   (3,10)


Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!


Thanks EP :)

Melody 22.02.2018
Melody 17.02.2018

1.A telegraph has x arms and each arm is capable of (x-1) distinct positions , including the position of rest. The total no. of signals that can be made is?

Rosala, This is just      x(x-1)



2. How many natural numbers are there fro 1 to 1000 which have none of their digits repeated?

i've tried this question a no. of times but it just doesnt see to be clicked in my mind. i would be really appreciate if anyone can explain it to me in detail.

There are 9 one digit ones

There are no four digit ones

So how many 2 digit ones are there.

The tens digit can be 1 to 9 that is 9 choices, now you used one digit but you can use the zeros so there are 9 possible digits left for the units digit

So that is 9*9=81

AND how many three digit ones are then  9*9*8 = 648

Altogether there are  9+81+648 = 738


3. Number of different natural numbers which are smaller than two hundred million and using only the digits 1 or 2 is :


less than      200,000,000   only containing the digits 1 and 2


1 digit     2

2 digit     2*2=2^2

3 digit      2^3

4 digit       2^4

5 digit      2^5

6 digit      2^6

7 digit      2^7

8 digit      2^8

9 digit      2^8       The biggest value digit must be 1


So what do we get when we add these up


\(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^8\\ =(2+2^2+2^3+2^4+2^5+2^6+2^7+2^8)+2^8\\ \qquad \text{The brackets is the sum of a GP, a=2 and r=2\;\;n=8}\\ =\frac{a(r^n-1)}{r-1}\;\;\;+2^8\\ =\frac{2(2^8-1)}{2-1}\;\;\;+2^8\\ =2^9-2+2^8\\ =2^9+2^8-2\\ =2^8(2+1)-2\\ =3\times 2^8-2 \)



i. (3).2^8 - 2 


ii. (3).2^8 - 1 


iii. 2 (2^9 - 1) 


iv) None 


I have a few more question like the questions above, and i just dont seem to get the correct answers. 


i have tried the questions above many times but i just cant understand a few things , i would be glad if anyone could explain to me in detail. Thank you very much. 


i really appreciate the time you'd take out for my answers! 

Melody 17.02.2018