Question

Solve the inequality :-
$({( \frac{1}{2}})^{x} - 2) \times \frac{ {x}^{2}(x + 1) }{(x + 2)} \leqslant 0$

Answer 1

Final Answer :
$x > - 2$

or x belongs to (-2,inf)

Steps and Understanding:
1)
$( { (\frac{1}{2} )}^{x} - 2) \times {x}^{2} \times \frac{(x + 1)}{(x + 2)} \leqslant 0 \\ = > (1 - {2}^{x + 1} ) \times {x}^{2} \times \frac{(x + 1)}{(x + 2)} \leqslant 2 \times 0 \\ = > ( {2}^{x + 1} - 1) \times {x}^{2} \times \frac{(x + 1)}{(x + 2)} \geqslant 0$

(2) Critical points of this equation are :

${2}^{x + 1} - 1 = 0 \\ = > {2}^{x + 1} = {2}^{0} \\ = > x + 1 = 0 = > x = - 1$
${x}^{2} = 0 = > x = 0 \: or \: 0 \\ (x + 1) = 0 = > x = - 1 \\ (x + 2) = 0 \\ = > x = - 2$

3) We got x = 0 ,-1 even number(two) of times, so there will be no sign change.
at x = -2, occurs only one time or odd time ,so there will be sign change.
See pic for sign and number line.

(4) Otherwise, we can check points in between the critical points to check sign.

(5) On seeing pic, we observe that for
Function we obtained after simplication is positive when
$x > - 2$
We are not using equality sign as 1/(x+2) will not be defined.
Hence, required value of x is x > -2.