i was doing some limits graphing and i notice this. I was doing a right hand limit. How come x>4 is 0>4-x and not x-4>0? the original function is limit x->4+ 3/(4-x)^3
x>4 is x-4>0
this approaches negative infinity since the top is poisitive and the bottom is negative and close to zero as x gets closer to 4
the guide line in the book says 4-x<0
x-4>0 -(x-4)<0 4-x<0
whenever you multiply or divide by negative the inequality flips
x > 4 is just like an equation except for the fact that it isnt "equal"; but the same steps apply in solving: x > 4 ; subtract x from both sides to get: 0 > 4-x ; if we flip this around the inequality flips as well: 4-x < 0 ; is the same statement
4>2 -4<-2
i mean how come its not x-4>0? i bring the 4 over instead or the x
it is the same
that is true x-4>0
-(x-4) does not equal x-4
0>4-x\[ and \not \neq\] x-4>0
oops 0>4-x≠ x-4>0
one says: 0 < 10; the other says -10 < 0
oh so its the same?
but when i graph it in the calculator it gives off a left hand limit
4-x > 0 is the same as saying: 0 < -(4-x) 0 < x-4
|x|>0 -x<0<x (x not equal to 0) |x-4|<0 -(x-4)<0<x-4 as long as x does not equal 4
i think you are losing your mind lol
could be :)
lol yeah i been zooom through the algebra to cal 1 review in a week
it would help to know what the original problem is asking for tho ;)
later guys gl losttime
thanks though the problem got me curious, the guideline showed x>4 is 0>4-x but when i was doing it i got x-4>0 and i got curious what if they brang the 4 over and the answer was different, but yeah i'll look it over
oh lmao right yes its the same the negative 1 will reverse the sign, silly me
but do u understand that is just removing one of the variables away and solving for the other?
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