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

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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)
 
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Since $$\left|\frac{1-x}{x}\right|=\frac{|1-x|}{|x|}$$, 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.
 
[math]f(x)= \frac{1- x}{x}= \frac{1}{x}- 1[/math] 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.
 
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Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
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