# Given that f(x)=x+1 and g(x)=1-x solve the inequality f(x)/g(x)>0

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### 3 Answers

We have the functions f(x)=x+1 and g(x)=1-x.

Now f(x) / g(x) > 0

=> (x +1)/(1-x) >0

This is true if either both x +1 and 1-x are greater than 0 or both are less than 0.

Taking the first case

x+ 1> 0 and 1- x > 0

=> x > -1 and 1 > x

So x lies in (1 , -1)

The second case gives

x +1 < 0 and 1 - x < 0

=> x < -1 and -x < -1

=> x< -1 and x > 1

This is not possible.

**Therefore x lies in (-1 , 1)**

f(x)=x+1 and g(x)=1-x. To solve the inequality f(x)/g(x)>0

Therefore to Let R(x) = f(x)/g(x) = (x+1)/(x-1) > 0.

R(x) = (x+1)/(x-1) > 0.

When x > 1, R(x) > 0, as both numerator x+1) > 0 and denominator x-1> 0 .

When when x = 1, R(x) = (x+1)/(x-1) is not defined as denominator x-1 = 0.

When -1 < x< 1, numerator is positive and denominator negative. So R(x) = (x+1)/(x-1) < 0.

When x= -1, both numerator = 0. Denominator = -2. So R(x) = (x+1)/(x-1) = 0.

When x< -1, both numerator and denominator are negative. So R(x) = (+1)/(x-1) > 0.

Therefore f(x)/g(x) = (x+1)/(x-1) > 0 .**only when x< -1 OR when x > 1.**

We'll substitute f(x) and g(x) by the given expressions:

(x+1)/(1 - x) > 0

Now, we'll discuss 2 cases:

1) The fraction is positive if both numerator and denominator are positive:

x + 1> 0

x > -1

1 - x > 0

-x > -1

x < 1

The common range of values that satisfy both inequalities is: (-1 ; 1).

2) The fraction is positive if both numerator and denominator are negative:

x + 1 < 0

x < -1

1 - x < 0

-x < -1

x > 1

In this case, there is no such an interval of values to satisfy both inequalitites.

**So, the interval of admissible values for the inequality f(x)/g(x)>0 to hold is (-1 ; 1).**