Melody

Does the last graph have the point (5,2) on it?

divide 9(cos5pi/4+i sin5pi/4)/3(cos pi+i sin pi)

\(9(cos5pi/4+i sin5pi/4)/3(cos pi+i sin pi)\\ =\frac{9e^{(5\pi/4)i}}{3e^{(\pi/4)i}}\\ =\frac{3e^{(5\pi/4)i}}{e^{(\pi/4)i}}\\ =3e^{(5\pi/4)i-(\pi/4)i}\\ =3e^{(4\pi/4)i}\\ =3e^{\pi i}\\ =3(cos\pi+i sin\pi)\\ =3(-1+0)\\ =-3\)

I think you mean

f(x)= Sqrt (x-1)  becasue none of them are   f(x)= Sqrt x-1

f(x)= Sqrt (x-1)

you know   Sqrt (4) = 2

and

f(x)= Sqrt (5-1)=sqrt(4)

so  lets think about  x=5 

f(5)= sqrt(5-1)  =   sqrt4  =  2

So (5,2) must be a point on the graph.

So which is the correct graph?

I discussed this question a day or two ago!!

Thanks Heureka,

I found my own error too :)

It was just a careless sign error in my last line.

\(x=log_2(({\frac{1-y}{y}})^{-1})\\ x=-log_2\left({\frac{y}{1-y}}\right) \qquad \text{This should have been } \color{red}{x=+log_2\left({\frac{y}{1-y}}\right) }\\ \therefore f^{-1}(x)=\color{red}{+} log_2\left({\frac{x}{1-x}}\right)\\ \text{Using the change of base rule this is exactly the same as Heureka's answer.} \)

Your answer from the internet is no doubt correct, it also matches Heureka's answer I think.

BUT

This is not correct       Can anyone spot what mistake I have made??

\(f(x)=\frac{2^x}{1+2^x}\\ let \\ y=\frac{2^x}{1+2^x}\\ \text{An initial inspection tells me that y>0}\\ \text{make x the subject}\\ y=\frac{2^x}{1+2^x}\\ \frac{1}{y}=\frac{1+2^x}{2^x}\\ \frac{1}{y}=\frac{1}{2^x}+\frac{2^x}{2^x}\\ \frac{1}{y}=\frac{1}{2^x}+1\\ \frac{1}{y}-1=2^{-x}\\ \frac{1-y}{y}=2^{-x}\\ log_2({\frac{1-y}{y}})=log_2(2^{-x})\\ log_2({\frac{1-y}{y}})=-x*log_2(2)\\ log_2({\frac{1-y}{y}})=-x\\ x=-log_2({\frac{1-y}{y}})\\ x=log_2(({\frac{1-y}{y}})^{-1})\\ x=-log_2\left({\frac{y}{1-y}}\right)\\ \therefore f^{-1}(x)=-log_2\left({\frac{x}{1-x}}\right)\\ \)

I have graphed this here

You have had very little response because your question does not make sense.

You are confusing similar with congruent.    (Congruent basically means 'the same')

Thes two shapes are similar because if you take the little one and dilate it (b**w it up) about its centre then you will get bigger circles.

If you dilate it with a factor of   s/r   then the radius will be s  and it will be congruent to the big circle. 

This is not a very technical answer but you are at an elementary level.  

Similar objects have the same basic shape but they can be bigger or smaller than each other. 

The angles will all be the same size and the sides will be in the same ratio.

I'd say 'false' but it is a very weirdly worded question.