Question

Solve for x Signum(x^3+x^2-6x)+1=0

Answer 1

Final Answer:
sgn(x^3+x^2-6x) = 0 if x = 0,-3,2
sgn( x^3+x^2-6x) = 1 if -3<x<0 or x>2
sgn(x^3+x^2-6x) =(-1 ) if. 0 <x<2 or x<-3


For Calculation and understanding see pic

Answer 2

If sgn (x³ – x² – 6x) + 1 = 0, then x ∈ (-∞, -3) ∪ (0, 2)

Step-by-step explanation:

Given

$\text{sgn} (x^3+x^2-6x)+1=0$

$\implies \text{sgn} (x^3+x^2-6x)=-1$  ..... (1)

We know that

$\sgn f(x)=\frac{|f(x)|}{f(x)}$

And

$|f(x)|=f(x)$ for all x for which $f(x)\ge0$

$|f(x)|=-f(x)$ for all x for which $f(x)<0$

Here

$f(x)=x^3+x^2-6x$

$\implies f(x)=x(x^2+x-6)$

$\implies f(x)=x(x^2+3x-2x-6)$

$\implies f(x)=x[x(x+3)-2(x+3)]$

$\implies f(x)=x(x+3)(x-2)$

Equation (1) will be satisfied for all x for which $x^3+x^2-6x<0$

Now,

$f(x)<0$

$\implies x(x+3)(x-2)<0$

$\implies x\in (-\infty,-3)\cup(0,2)$

Hope this answer is helpful.

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