Is this the correct way to do this?

[tired bluff]

i think my work is wrong but i still got the right answer anyway

assuming m+1/(m-1) >0 is wrong i think but idk what would be the correct way to do this

[broken barn]

you still haven't set the range

they just want to know the range of m

you have found one awnser

m > 1

[hot mantle]

Isn't that a suitable 'range' for m

[digital solar]

I assume y has to be real for all x.
In particular take x = 2m. What can you say then?

[livid crag]

yeah its only that right?

(1,inf)

[livid crag]

how did you get( x-2m)^2>m?

i got (x-2m)^2+m+1/(m-1)>0

just find minimum of that

you get a quadratic in m

graph thr numerator

then solve the denominator part

[digital solar]

Is x fixed or variable?

[digital solar]

If variable then ^

[tired bluff]

@livid crag the minimum of (x-2m)^2 is 0 so would you just get 0+m+1/(m-1)>0 and do what i did there?

[fleet cypress]

just try to prove that it has imaginary roots

which means the quadratic is bounded to above or below x-axis

for which you'll get m + 1/m-1 > 0

you can easily get the range of m then

it just needs to avoid [1,-1]

[tired bluff]

is this right?

i used a discriminant trick to find range of -m - 1/(m-1)

[fleet cypress]

and put a value to see where it ends only positive

or negative

since if you put m < -1, you'll only get positive values for the f(x) for any x

m > 1

[tired bluff]

the answer key says m>1 is correct but in the my first attempt my work is wrong but somehow still got the right answer

[fleet cypress]

you're overcomplicating it

you just to find the values of m where f(x) has no real zeros

one case for the upper quadrants and one case for the lower quadrants

since log_3 isn't defined for negative numbers yeet the range that gives the lower quadrant range

[tired bluff]

.close thx guys