Question

Por que x² ≤ 4 = -2 ≤ x ≤ 2 ?

De onde surge a parte do (- 2)? alguém me explica?

Por que não x ≤ 2?

Answer 1

$\Large\boxed{\begin{array}{l}\underline{\rm Observe\,a\,figura\,que\,eu\,anexei.}\\\sf x^2\leqslant4\\\sf x^2-4\leqslant0\\\underline{\boldsymbol{fac_{\!\!,}a}}\\\sf f(x)=x^2-4\\\underline{\rm zeros\,de\,f(x):}\\\sf x^2-4=0\\\sf x^2=4\\\sf x=\pm\sqrt{4}\\\sf x=\pm2\\\sf A\,par\acute abola\,intercepta\,o\,eixo\,x\\\sf nos\,pontos\,A(-2,0)\,e\,B(2,0).\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm Estudo\,do\,sinal\,de\,f(x):}\\\sf f(x)=x^2-4\\\sf a=1>0\longrightarrow concavidade\,para\,cima\\\sf f(x)\leqslant0\iff -2\leqslant x\leqslant 2\\\sf note\,que\,se\, dizermos\\\sf simplesmente\,que\,x\leqslant2\,pode\,n\tilde ao\, ficar\,claro\\\sf se\,de\,fato\,qualquer\,valor\,de\,x\,inferior\,a\,2\\\sf satisfaz\,a\,inequac_{\!\!,}\tilde ao\,dada.\\\sf note\,que\,x=-3<2\\\sf a\,inequac_{\!\!,}\tilde ao\,n\tilde ao\,\acute e\,satisfeita\end{array}}$

$\Large\boxed{\begin{array}{l}\sf pois\,(-3)^2=9>4\\\sf e\,queremos\,valores\,de\,x\,tais\,que\, x^2\leqslant4.\\\sf Por\,esta\,raz\tilde ao\\\sf precisamos\,ser\,categ\acute oricos\,na\,hora\,de\,exibir\\\sf o\,intervalo\,que\,de\,fato\,soluciona\,a\,inequac_{\!\!,}\tilde ao.\end{array}}$