Question

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

How to make x the subject of the formula??

Answer 1

You mean you want to solve for x ?

Answer 2

in terms of y?

Answer 3

yes.

Answer 4

ok so first of all :
\[\large{y=\frac{x^2+x+1}{x}}\]
\[\large{y*x=x^2+x+1}\]

Answer 5

"the subject of the formula" what ever happened to "solve for \(x\)"?

Answer 6

\[\large{xy=x^2+x+1}\]
\[\large{xy-x^2-x=1}\]
\[\large{x(y-x-1)=1}\]
\[\large{x=\frac{1}{y-x-1}}\]

Answer 7

hmmm

Answer 8

well. that doesn't help. ALL the x terms need to be on one side :P

Answer 9

I think this is called the inverse of the function.

Answer 10

you need to solve a quadratic equation

Answer 11

this is not a one to one function, so doesn't have an inverse, but we can still solve for \(x\)

Answer 12

After you get here you wanna put everything on one side and use the quad formula
Have fun (hint: group your x terms together)
\[\large{xy-x^2-x=1}\]

Answer 13

\[\large{xy=x^2+x+1}\]
\[x^2+x-xy+1=0\]
\[x^2+(1-y)x+1=0\] now use the quadatic formula with \(a=1,b=1-y, c=1\) to solve for \(x\)

Answer 14

See, i don't really need to solve it as much as find an equation for the so called inverse, so that i can find it's domain. the whole point of this is to find the range of the original equation, and to do that, i find the domain of its inverse. if you have any other method, please enlighten me :P

Answer 15

Oh Like Sat did.

Answer 16

yes like i did so you don't have all those pesky negative coefficients!

Answer 17

\[\large{x=\frac{1}{y-x-1}}\]
\[\large{xy-x^2-x=1}\]
\[\large{xy-x^2-x-1=0}\]
\[\large{-x^2+x(y-1)-1=0}\]
\[\large{x=\frac{1-y\pm \sqrt{(y-1)^2-4}}{-2}}\]
\[\large{x=\frac{1-y\pm \sqrt{y^2+1-2y-4}}{-2}}\]

Answer 18

no this is the method you need so use
solve for \(x\) and then note that the discrinant (the expression inside the radical) must be greater than or equal to zero
solve that to get your range

Answer 19

*discriminant

Answer 20

why greater than or equal to 0, though?? it has 2 distinct real roots??

Answer 21

you cannot take the square root of negative number

Answer 22

DUH.

Answer 23

so in order for \(x\) to be a real number, it must be the case that the expression inside the radical has to be non-negative

Answer 24

OH.

Answer 25

assuming that there is no mistake in the work done by mathlover above, the expression inside the radical is \((y-1)^2-4\) so you have to solve
\((y-1)^2-4\geq 0\) for \(y\)
this will give you the range

Answer 26

away from this topic, though, for one sec, did you know you had 1287 fans??? THATS INSANE.

Answer 27

yes, many of them are from that category i assume

Answer 28

But anywayss, THANK YOU so muchh :D

Answer 29

i hope it is clear why this works
you are looking for the range of
\[f(x)=\frac{x^2+x+1}{x}\] so you find the "inverse" and this means what? you pick a \(y\) and say "find the \(x\) that produces it
if you pick a \(y\) with \((y-1)^2-4<0\) there will be no such \(x\) whereas if you pick a \(y\) with
\[(y-1)^2-4\geq 0\] there will be such an \(x\)
that is why this will give the range

Answer 30

a quick check shows @amistre64 has more fans than i do, leading me to believe that his asylum base is even larger than mine

Answer 31

i have fans?

Answer 32

paper ones, imported from taiwan

Answer 33

was you around when fan count was the equivalent of a "medal" count/smartscore?

Answer 34

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