Find all values of x that satisfy the inequality.
x^2-16
______ <(or equal to) 0
x^2-1
2007-01-05 09:12:47 · 4 answers · asked by Sar 3 in Science & Mathematics ➔ Mathematics
x^2-16/x^2-1<0
x^2-16<0
x^2<16
x>-4 or x<4
if x^2-16/x^2-1=0
x^2-16=0
x^2=16
x=+/-4
sounder thecircumstances when
x^2-16/x^2-1<=0
x<=-4 and x>=4
2007-01-05 09:18:31 · answer #1 · answered by raj 7 · 1⤊ 1⤋
x has a range from -4 to 4
excluding 1 n -1
as at these pts eqn tends to infinity
as x^-16<=0gives range
2007-01-05 17:22:55 · answer #2 · answered by well thts it...... 3 · 0⤊ 0⤋
> solve((x^2-16)/(x^2-1)<0,x);
RealRange(Open(-4), Open(-1)), RealRange(Open(1), Open(4))
So the open interval (-4,-1) union the open interval (1,4)
Hey raj limit x -> +/- 00 is 1 which is not <0 so you better re-think this one.
2007-01-05 17:21:24 · answer #3 · answered by a_math_guy 5 · 1⤊ 0⤋
-4 (lessthanorequalto)X(lessthanorequalto) -1
and
4 (greatrthanoequal) X (greatrthanoequal) 1
2007-01-05 22:34:36 · answer #4 · answered by gregory_s19 3 · 0⤊ 0⤋