Question

PLEASE HELP

Answer 1

$\\ \sf\longmapsto \dfrac{x^2-1}{x^2+5x+4}\leqslant 0$

• Solving denominator

$\\ \sf\longmapsto x^2+5x+4>0$

$\\ \sf\longmapsto x^2+4x+x+4>0$

$\\ \sf\longmapsto x(x+4)+1(x+4)>0$

$\\ \sf\longmapsto (x+4)(x+1)>0$

$\\ \sf\longmapsto x>-4\:or x>-1$

• Hence x$\neq$-1
• x can't be 0 as it makes function undefined

$\\ \sf\longmapsto -4$