Question

Determine o conjunto solução da inequação 2 ^(x + 2) + 2^(-1-x) < 3

Answer 1

Resposta:

-2 < x < -1

Explicação passo-a-passo:

2⁽ˣ⁺²⁾+2⁽⁻¹⁻ˣ⁾<3

2ˣ.2²+2⁻¹.2⁻ˣ<3

4.2ˣ+1/(2.2ˣ)<3

(2.2ˣ).4.2ˣ+1<3.(2.2ˣ)

8.2²ˣ+1<6.2ˣ

8.2²ˣ-6.2ˣ+1<0

Chamando y=2ˣ

8y²-6y+1<0

$\displaystyle Aplicando~a~f\'{o}rmula~de~Bhaskara~para~8y^{2}-6y+1=0~~e~comparando~com~(a)y^{2}+(b)y+(c)=0,~determinamos~os~coeficientes:~\\a=8{;}~b=-6~e~c=1\\C\'alculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(-6)^{2}-4(8)(1)=36-(32)=4\\C\'alculo~das~raizes:&\\y^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(-6)-\sqrt{4}}{2(8)}=\frac{6-2}{16}=\frac{4\div4}{16\div4}=\frac{1}{4}\\\\y^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(-6)+\sqrt{4}}{2(8)}=\frac{6+2}{16}=\frac{8\div8}{16\div8}=\frac{1}{2}$1/4<y<1/2, mas y=2ˣ

1/4<2ˣ<1/2

4⁻¹<2ˣ<2⁻¹

2²⁽⁻¹⁾<2ˣ<2⁻¹

2⁻²<2ˣ<2⁻¹

-2 < x < -1