## mathmath333 one year ago Find the range of $$x$$.

1. mathmath333

\large \color{black}{\begin{align} \dfrac{(x+3)(x^2-4)}{(x-1)^2}\leq 0\hspace{.33em}\\~\\ \end{align}}

2. geerky42

Since $$(x-1)^2$$ is always positive, Multiply both sides by that, then we are down to $$(x+3)(x^2-4)\le0$$

3. geerky42

Note that $$x\neq1$$

4. imqwerty

1<x<=2

5. geerky42

Then try find root of $$(x+3)(x^2-4)=0$$, then determine which intervals between roots exclusive would have $$(x+3)(x^2-4)$$ be smaller than 0

6. geerky42

Yeah but there are more @imqwerty

7. idku

at first $$x\ne1$$ now, for this right side to be =0, x=2, -2, -3 for the right side to be greater than 0, (x+3) and (x^2-4) , both have to be negative, or both positive NOTE: (x-1)^2 is always positive x+3 and (x-2)(x+2) are both pos. when x>-2 x+3 and (x-2)(x+2) are both neg. when x<-3 x>-2 or x<-3 and x = 2, -2, -3 CONCLUSION x$$\le$$-3, or x $$\ge$$ -2 and x=2

8. campbell_st

well if you multiply both sides of the equation by (x -1)^2 then you get $(x + 3)(x^2-4) \le 0$ so this becomes $(x + 3)(x-2)(x+2) \le 0$ so the zeros are x= -3, -2 and 2 so pick some test values that are between the zeros and check |dw:1433794059127:dw| x = 0 you get -12/1 pick x = -2.5 you get a positive

9. campbell_st

so the graph would look like |dw:1433794220437:dw| in some way... hope it makes sense

10. campbell_st

now you can define the intevals where the inequality holds true.

11. campbell_st

there is an oblique asymptote which won't affect the solution

12. imqwerty

x<=3 or -2<=x<1 or 1<x<=2

13. geerky42

14. campbell_st

15. imqwerty

:)

16. campbell_st

but then again @imqwerty I think you solution is still incorrect... Can't see how 1 plays a roll in the solution

17. imqwerty

1 can't be the solution because then the denominator will become 0 @campbell_st

18. campbell_st

good point