Hello, psolaki!
There are two "critical values": .
The two values divide the number line into three intervals.
. .
Test a value of in each interval.
The inequality is true on the interval
Answer: .
in trying to solve the following inequality : the following proof went through my mind ,but i am not sure if it is correct:
Proof:
Assume 1-x<0 ,then the inequality becomes ,if we multiply across : 1/x <0 ,which implies that x<0.
BUT since 1<x ( 1-x<0) we have 1<0 , a contradiction,hence .
BUT since [(1-x)<0 or (1-x)>0] ,and since we have (1-x)>0
Hence the inequality is satisfied for all ,x<1
Thank you Soroban,but can you help me ,and tell me where my solution is wrong??
Because i have the following dilemma.
Suppose we wanted to solve out the inequality:
And suppose that we followed the same argument that i used for solving the inequality in my opening post:
proof:
Assume x>o,then the inequality becomes,if we multiply across: ,which implies that ,a contradiction ,hence .
BUT since [x<0 or x>0] ,and since we have x<o
Hence the inequality is true for all x<0.
So how can the same TYPE of argument in one case give the right solution and the wrong solution in another case??
Your logic is wrong. Yes, because x can't be 0 on the left, we must have x> 0 or x< 0. Since x> 0 does not satisfy the inequality, we must have x< 0.
NOT necessarily all x< 0.Hence the inequality is true for all x<0.
You haven't used the full argument.So how can the same TYPE of argument in one case give the right solution and the wrong solution in another case??
if x< 0, multiplying by x reverses the inequality:]
Subtracting x+ 1 from both sides
In order that the product of two numbers be negative the two numbers must have opposite signs. Either
x-1> 0 and x+ 1< 0 or x-1< 0 and x+ 1> 0.
x- 1> 0 gives x>1 but remember that we are assuming x< 0 so this is not a valid solution.
x-1< 0 gives x< 1 but since x< 0, that gives nothing new. x+1> 0 gives x> -1. -1< x< 0 satisfies both conditions.
The solution set to the inequality is -1< x< 0.