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Question Number 210496 by lmcp1203 last updated on 11/Aug/24

Commented by mr W last updated on 11/Aug/24

$h a s t h a t b r a c k e t a s p e c i a l m e a n i n g ?$
Commented by lmcp1203 last updated on 11/Aug/24

$p l e a s e h e l p m e t h a n k s$
Commented by klipto last updated on 11/Aug/24

$that should be floor function$
Commented by lmcp1203 last updated on 11/Aug/24

$y e s$
	Answered by Frix last updated on 11/Aug/24
	$f(x)=∣3−2x∣= { ((3−2x, x≤(3/2))),((2x−3, x>(3/2))) :} g(x)=⌊∣−3x∣⌋+⌊4x+1⌋= = { ((1+⌊−3x⌋+⌊4x⌋, x≤0)),((1+⌊3x⌋+⌊4x⌋, x>0)) :} ⇒ f(x)≥0, x∈R f((3/2))=0 g(x)≤0, x<0; g(x)≥1; x≥0 ⇒ possible solutions for x∈]0, (3/2)[ ⇒ 3−2x=1+⌊3x⌋+⌊4x⌋ h(x)=2x+⌊3x⌋+⌊4x⌋=2 lim_(x→1/3^− ) h(x) =−(1/3) lim_(x→1/3^+ ) h(x) =+(2/3) ⇒ no solution$
	$f (x) =∣ 3 - 2 x ∣= {\begin{cases} 3 - 2 x, x ⩽ \frac{3}{2} \\ 2 x - 3, x > \frac{3}{2} \end{cases}$ $g (x) = ⌊ ∣ - 3 x ∣ ⌋ + ⌊ 4 x + 1 ⌋ =$ $= {\begin{cases} 1 + ⌊ - 3 x ⌋ + ⌊ 4 x ⌋, x ⩽ 0 \\ 1 + ⌊ 3 x ⌋ + ⌊ 4 x ⌋, x > 0 \end{cases}$ $\Rightarrow$ $f (x) ⩾ 0, x \in R f (\frac{3}{2}) = 0$ $g (x) ⩽ 0, x < 0; g (x) ⩾ 1; x ⩾ 0$ $\Rightarrow possible solutions for x \in] 0, \frac{3}{2} [$ $\Rightarrow$ $3 - 2 x = 1 + ⌊ 3 x ⌋ + ⌊ 4 x ⌋$ $h (x) = 2 x + ⌊ 3 x ⌋ + ⌊ 4 x ⌋ = 2$ $\lim_{x \to 1 / 3^{-}} h (x) = - \frac{1}{3}$ $\lim_{x \to 1 / 3^{+}} h (x) = + \frac{2}{3}$ $\Rightarrow no solution$
	Commented by lmcp1203 last updated on 12/Aug/24

	$t h a n k s s i r$
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