Solving g(x)-h(x) <0: Find a, b, c, d

[Manal]

We have :
f(x) =(1-x) ÷x, x is in IR*
g(x)=|f(x)|
h(x) =4-x^2
Solve g(x)-h(x) <0 (not with a graph)
( I tried solving so , S=]a, b[ U ]c,d[., but i don't know what are a, b, c and d. I would appreciate the help a bunch)

 

[Evgeny.Makarov]

Since \(\displaystyle \left|\frac{1-x}{x}\right|=\frac{|1-x|}{|x|}\), absolute values can be eliminated on each of the intervals $(-\infty,0)$, $(0,1)$ and $[1,\infty)$ where $1-x$ and $x$ have definite signs. On each interval you get a cubic inequality.

 

[HOI]

\(\displaystyle f(x)= \frac{1- x}{x}= \frac{1}{x}- 1\) is 0 at x= 1 and is not continuous at x= 0. Those are the only places f can change sign. If x= -1, f(-1)= -2 so f(x) is negative for all x less than 0. If x= 1/2 f(1/2)= 2- 1= 1 so f(x) is positive for all x between 0 and 1. Finally, if x= 2, f(x)= 1/2- 1= -1/2 so x is negative for all x greater than 1.

g(x)= |f(x)|= 1- 1/x for x< 0
g(x)= 1/x- 1 for 0< x< 1
g(x)= 1- 1/x for 1< x.

If x< 0, g(x)- h(x)= 1- 1/x- 4- x^2= -x^2- 1/x- 3 for x< 0 so you want to solve -x^2- 1/x- 3< 0. Multiplying by -x, which is positive, x^3+ 3x+ 1< 0.

If 0< x< 1, g(x)- h(x)= 1/x- 1- 4- x^2= -x^2+ 1/x- 5 so you want to solve -x^2+ 1/x- 5< 0. Multiplying by -x, which is negative, x^3+ 5x- 1> 0.

If x> 1, g(x)- h(x)= 1-1/x- 4- x^2= -x^2- 1/x- 3 so you want to solve -x^2- 1/x- 3<0. Multiplying by -x, which is negative, x^3+ 3x+ 1> 0.