If, f(x)=− x^2 +4x −3 , g(x)= { (( (√(7−x)) 0≤x <7)),(( ⌊ 5x ⌋ −5x x≥7)) :} ⇒ R_(fog) = (a ,b] find the value of b−a R_(fog) = { (fog)(x)∣ x∈ D


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Question Number 212140 by mnjuly1970 last updated on 03/Oct/24
$If, f(x)=− x^2 +4x −3 , g(x)= { (( (√(7−x)) 0≤x <7)),(( ⌊ 5x ⌋ −5x x≥7)) :} ⇒ R_(fog) = (a ,b] find the value of b−a R_(fog) = { (fog)(x)∣ x∈ D_(fog) }$
$I f, f (x) = - x^{2} + 4 x - 3$ $, g (x) = {\begin{cases} \sqrt{7 - x} 0 ⩽ x < 7 \\ ⌊ 5 x ⌋ - 5 x x ⩾ 7 \end{cases}$ $\Rightarrow R_{f o g} = (a, b]$ $f i n d t h e v a l u e o f b - a$ $R_{f o g} = {(f o g) (x) ∣ x \in D_{f o g}}$
	Answered by mehdee7396 last updated on 04/Oct/24

	$R_{g} = (- 1, 0] \cup (0, \sqrt{7}] = (- 1, \sqrt{7}]$ $f (x) = - {(x - 2)}^{2} + 1$ $x \in D_{g} \Rightarrow - 1 < x ⩽ \sqrt{7} \Rightarrow - 3 < x - 2 ⩽ - 2 + \sqrt{7}$ $0 ⩽ {(x - 2)}^{2} < 9$ $- 9 < - {(x - 2)}^{2} ⩽ 0$ $- 8 < R_{f o g} ⩽ 1$ $\Rightarrow b - a = 9 ✓$
	Commented by mnjuly1970 last updated on 04/Oct/24

	$t h a b k s a l o t s i r m e h d e e$
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