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Keith_Richards [23]
3 years ago
7

PLEASE HELP

Mathematics
1 answer:
kifflom [539]3 years ago
5 0

\\ \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

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