f(x)= (( x +m)/(∣x∣ + 6)) and f is strictly monoton .find the value(s) of m . m∈Z


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Question Number 165552 by mnjuly1970 last updated on 03/Feb/22

$f (x) = \frac{x + m}{∣ x ∣ + 6} a n d f i s$ $s t r i c t l y m o n o t o n . f i n d t h e v a l u e (s)$ $o f m . m \in Z$
	Answered by TheSupreme last updated on 03/Feb/22

	$f^{'} (x) > 0$ $\frac{∣ x ∣ + m - s g n (x) (x + m)}{{(∣ x ∣ + 6)}^{2}} = \frac{∣ x ∣ + m - ∣ x ∣ - s g n (x) m}{{(∣ x ∣ + 6)}^{2}} > 0 \to m (1 - s g n (x)) > 0$ $∄ m ∣ \forall x \in R : f (x) i s s t r i c t l y m o n o t o n$ $f (x) i s m o n o t o n f o r m = 0 \to f (x) = s g n (x)$
	Answered by mahdipoor last updated on 03/Feb/22
	$⇒f= { ((((x+m)/(x+6)) x>0)),((((x+m)/(6−x)) x≤0)) :} ⇒f^′ = { ((((6−m)/((x+6)^2 )) x<0)),((((m+6)/((6−x)^2 )) x≥0)) :} { ((6−m<0 & m+6<0 ⇒ ∄)),((6−m>0 & m+6>0 ⇒ −6<m<6)) :} m∈Z⇒m∈{−5,−4,...,4,5}$
	$\Rightarrow f = {\begin{cases} \frac{x + m}{x + 6} x > 0 \\ \frac{x + m}{6 - x} x ⩽ 0 \end{cases} \Rightarrow f^{^{'}} = {\begin{cases} \frac{6 - m}{{(x + 6)}^{2}} x < 0 \\ \frac{m + 6}{{(6 - x)}^{2}} x ⩾ 0 \end{cases}$ ${\begin{cases} 6 - m < 0 & m + 6 < 0 \Rightarrow ∄ \\ 6 - m > 0 & m + 6 > 0 \Rightarrow - 6 < m < 6 \end{cases}$ $m \in Z \Rightarrow m \in {- 5, - 4, . . ., 4, 5}$
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