Question

Please answer my questions no.11 and 15(3)..
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Answer 1

ANSWER....

$\frac{1}{ \sqrt{ |x| - x} } \\ \\ we \: know \: that \: square \: root \: can \\ never \: give \: a \: negative \: number... \\ \\ so \: \sqrt{ |x| - x} > 0 \\ \\ please \: note \: that \: in \: this \: case \: \\ square \: root \: is \: not \: equal \: to \: zero \\ \: because \: this \: give \: a \: indeterminant \\ \\ \sqrt{ |x| - x } > 0 \\ squaring \: both \: side \\ \\ |x| - x > 0 \\ \\ case \: 1 \\ \\ when \: x \: is \: positive \: number \\ \\ |x| - x > 0 \\ \\ x - x > 0 \\ \\ 0 > 0 \: (noth \: possible) \\ ( |x| = x \: if \: x \: is \: a \: positive \: number) \\ \\ case \: two \: when \: x \: is \: a \: negative \\ number \\ \\ |x| - x > 0 \\ \\ - x - x > 0 \\ \\ - 2x > 0 \\ \\ x < 0....all \: negative \: number \\ \\ |x| = - x \: if \: x \: is \: a \: negative \: number \\ \\ domain = all \: negative \: number \\ \\ \frac{1}{ \sqrt{ |x| - x} } \: as \: we \: know \: this \: can \: never \: be \: a \\ \\ \: negative \: number \: so \: range \: is \: \\ all \: positive$

NEXT PART....

$5 - |x + 1| \\ \\ as \: you \: see \: in \: the \: given \: question \: \\ there \: is \: no \: square \: root \: no \: denominetor \\ so \: we \: can \: put \: any \: no. \: in \: the \: place \: of \: x \\ \\ so \: domain \: is \: all \: real \: number \\ \\ range \: is \: all \: real \: number$

#BTSKINGDOM