Question

Find the range of \(x\).

Answer 1

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

Answer 2

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

Answer 3

Note that \(x\neq1\)

Answer 4

1<x<=2

Answer 5

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

Answer 6

Yeah but there are more @imqwerty

Answer 7

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

Answer 8

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




x = 0 you get -12/1

pick x = -2.5

you get a positive

Answer 9

so the graph would look like in some way...

hope it makes sense