Question find the” minimum” value of: f (x):= ∣1+x∣+∣ 2+x∣ + ∣4 +2x∣


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Question Number 157604 by mnjuly1970 last updated on 25/Oct/21

$Q u e s t i o n$ $f i n d {t h e}^{″} {m i n i m u m}^{″} v a l u e o f :$ $f (x) := ∣ 1 + x ∣ + ∣ 2 + x ∣ + ∣ 4 + 2 x ∣$
Commented by mr W last updated on 25/Oct/21

$i t h i n k t h e e a s i e s t w a y i s t o c h e c k a l l$ $k i n k p o s i t i o n s :$ $x = - 1, - 2$ $f (- 1) = 0 + 1 + 2 = 3$ $f (- 2) = 1 + 0 + 0 = 1$ $\Rightarrow f {(x)}_{m i n} = 1$
Commented by mr W last updated on 25/Oct/21
$in this case we can also do like this: f(x)=∣1+x∣+3∣1+1+x∣=∣1+x∣+3±3∣1+x∣ =3+ { ((4∣1+x∣)),((−2∣1+x∣)) :} f(x)_(min) =3−2=1 at x=−1$
$i n t h i s c a s e w e c a n a l s o d o l i k e t h i s :$ $f (x) =∣ 1 + x ∣ + 3 ∣ 1 + 1 + x ∣=∣ 1 + x ∣ + 3 \pm 3 ∣ 1 + x ∣$ $= 3 + {\begin{cases} 4 ∣ 1 + x ∣ \\ - 2 ∣ 1 + x ∣ \end{cases}$ $f {(x)}_{m i n} = 3 - 2 = 1 a t x = - 1$
Commented by mnjuly1970 last updated on 25/Oct/21

$g r a t e f u l s i r W$
	Answered by TheSupreme last updated on 25/Oct/21
	$f(x)=∣1+x∣+3∣2+x∣ f′(x)=sgn(1+x)+3sgn(2+x)≠0 ∀x∈R−{−1,−2} f(−1)=3 f(−2)=1 lim_(x→±∞) f(x)=∞ min f(x)=1$
	$f (x) =∣ 1 + x ∣ + 3 ∣ 2 + x ∣$ $f^{'} (x) = s g n (1 + x) + 3 s g n (2 + x) \neq 0 \forall x \in R - {- 1, - 2}$ $f (- 1) = 3$ $f (- 2) = 1$ $l i \underset{x \to \pm \infty}{m} f (x) = \infty$ $m i n f (x) = 1$
	Commented by mnjuly1970 last updated on 25/Oct/21

	$t h a n k s a l o t$
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