let f(x)=x−(√(4−x^2 )) and g(x) =((2 +(√(x−3)))/(2−(√(x−3)))) 1) find D_f ,D_g and D_(fog) and determine fog(x) 2) calculate gof(x) and give D_(gof) 3) calculate ∫_(−(1/2)) ^(1/2) f(x)dx 4) calculate ∫


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Question Number 59188 by maxmathsup by imad last updated on 05/May/19

$l e t f (x) = x - \sqrt{4 - x^{2}} a n d g (x) = \frac{2 + \sqrt{x - 3}}{2 - \sqrt{x - 3}}$ $1) f i n d D_{f}, D_{g} a n d D_{f o g} a n d d e t e r m i n e f o g (x)$ $2) c a l c u l a t e g o f (x) a n d g i v e D_{g o f}$ $3) c a l c u l a t e \int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x$ $4) c a l c u l a t e \int_{4}^{5} g (x) d x .$
Commented by maxmathsup by imad last updated on 06/May/19

$1) x \in D_{f} \Leftrightarrow 4 - x^{2} ⩾ 0 \Leftrightarrow x^{2} ⩽ 4 \Leftrightarrow∣ x ∣⩽ 2 \Leftrightarrow - 2 ⩽ x ⩽ 2 \Rightarrow D_{f} = [- 2, 2]$ $x \in D_{g} \Leftrightarrow x - 3 ⩾ 0 a n d 2 - \sqrt{x - 3} \neq 0 \Leftrightarrow x ⩾ 3 a n d \sqrt{x - 3} \neq 2 \Leftrightarrow x ⩾ 3 a n d x \neq 7 \Rightarrow$ $D_{g} = [3, 7 [\cup] 7, + \infty [$ $x \in f o g \Leftrightarrow x \in D_{g} a n d g (x) \in D_{f} \Leftrightarrow x ⩾ 3 a n d x \neq 7 a n d - 2 ⩽ g (x) ⩽ 2 b u t$ $- 2 ⩽ g (x) ⩽ 2 \Rightarrow 4 - g^{2} (x) ⩾ 0 \Rightarrow 4 - {(\frac{2 + \sqrt{x - 3}}{2 - \sqrt{x - 3}})}^{2} ⩾ 0 \Rightarrow$ $\frac{4 {(2 - \sqrt{x - 3})}^{2} - {(2 + \sqrt{x - 3})}^{2}}{{(2 - \sqrt{x - 3})}^{2}} ⩾ 0 \Rightarrow 4 (4 - 4 \sqrt{x - 3} + x - 3) - (4 + 4 \sqrt{x - 3} + x - 3) ⩾ 0$ $\Rightarrow 16 - 16 \sqrt{x - 3} + 4 x - 12 - 4 - 4 \sqrt{x - 3} - x + 3 ⩾ 0 \Rightarrow$ $- 20 \sqrt{x - 3} + 3 x + 3 ⩾ 0 \Rightarrow 3 x + 3 ⩾ 20 \sqrt{x - 3} \Rightarrow {(3 x + 3)}^{2} ⩾ 400 (x - 3) \Rightarrow$ $9 (x^{2} + 2 x + 1) ⩾ 400 x - 1200 \Rightarrow$ $9 x^{2} + 18 x - 400 x + 9 + 1200 ⩾ 0 \Rightarrow 9 x^{2} - 382 x + 1209 ⩾ 0$ $Δ^{^{'}} = {(191)}^{2} - 9.1209 = . . . .$
Commented by maxmathsup by imad last updated on 06/May/19

$f o g (x) = f (g (x)) = g (x) - \sqrt{4 - {(g (x))}^{2}} = \frac{2 + \sqrt{x - 3}}{2 - \sqrt{x - 3}} - \sqrt{4 - {(\frac{2 + \sqrt{x - 3}}{2 - \sqrt{x - 3}})}^{2}}$ $2) g o f (x) = g (f (x)) = \frac{2 + \sqrt{f (x) - 3}}{2 - \sqrt{f (x) - 3}} = \frac{2 - \sqrt{x - \sqrt{4 - x^{2}} - 3}}{2 - \sqrt{x - \sqrt{4 - x^{2}} - 3}}$ $x \in D_{g 0 f} \Leftrightarrow x - 3 - \sqrt{4 - x^{2}} ⩾ 0 a n d 4 - x^{2} ⩾ 0 a n d \sqrt{x - 3 - \sqrt{4 - x^{2}}} \neq 2 \Rightarrow$ $b u t 4 - x^{2} ⩾ 0 \Rightarrow - 2 ⩽ x ⩽ 2$ $x - 3 - \sqrt{4 - x^{2}} ⩾ 0 \Rightarrow x - 3 ⩾ \sqrt{4 - x^{2}} w e m u s t h a v e x ⩾ 3 b u t x \in [- 2, 2]$ $c o n d i t i o n i m p o s s i b l e s o g o f i s n o t d e f i n e d . . .!$
Commented by maxmathsup by imad last updated on 06/May/19
$3) ∫_(−(1/2)) ^(1/2) f(x)dx =∫_(−(1/2)) ^(1/2) (x−(√(4−x^2 )))dx =∫_(−(1/2)) ^(1/2) xdx −2∫_0 ^(1/2) (√(4−x^2 ))dx =0−2∫_0 ^(1/2) (√(4−x^2 ))dx =_(x=2sint) −2 ∫_0 ^(arcsin((1/4))) 2(√(1−sin^2 t))(2cost)dt =−8 ∫_0 ^(arcsin((1/4))) cos^2 t dt =−4 ∫_0 ^(arcsin((1/4))) (1+cos(2t))dt =−4 arcsin((1/4)) −2 [sin(2t)]_0 ^(arcsin((1/4))) =−4 arcsin((1/4))−2{ sin(2arsin((1/4))} but sin(2arcsin((1/4)))=2sin(arcsin((1/4)))(√(1−sin^2 (arsin((1/4))))) =(1/2)(√(1−((1/4))^2 ))=(1/2)(√(1−(1/(16))))=(1/2) ((√(15))/4) =((√(15))/8) ⇒ ∫_(−(1/2)) ^(1/2) f(x)dx =−4 arcsin((1/4))−((√(15))/4)$
$3) \int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x = \int_{- \frac{1}{2}}^{\frac{1}{2}} (x - \sqrt{4 - x^{2}}) d x = \int_{- \frac{1}{2}}^{\frac{1}{2}} x d x - 2 \int_{0}^{\frac{1}{2}} \sqrt{4 - x^{2}} d x$ $= 0 - 2 \int_{0}^{\frac{1}{2}} \sqrt{4 - x^{2}} d x =_{x = 2 s i n t} - 2 \int_{0}^{a r c s i n (\frac{1}{4})} 2 \sqrt{1 - {s i n}^{2} t} (2 c o s t) d t$ $= - 8 \int_{0}^{a r c s i n (\frac{1}{4})} {c o s}^{2} t d t = - 4 \int_{0}^{a r c s i n (\frac{1}{4})} (1 + c o s (2 t)) d t$ $= - 4 a r c s i n (\frac{1}{4}) - 2 {[s i n (2 t)]}_{0}^{a r c s i n (\frac{1}{4})}$ $= - 4 a r c s i n (\frac{1}{4}) - 2 {s i n (2 a r s i n (\frac{1}{4})} b u t$ $s i n (2 a r c s i n (\frac{1}{4})) = 2 s i n (a r c s i n (\frac{1}{4})) \sqrt{1 - {s i n}^{2} (a r s i n (\frac{1}{4}))}$ $= \frac{1}{2} \sqrt{1 - {(\frac{1}{4})}^{2}} = \frac{1}{2} \sqrt{1 - \frac{1}{16}} = \frac{1}{2} \frac{\sqrt{15}}{4} = \frac{\sqrt{15}}{8} \Rightarrow$ $\int_{- \frac{1}{2}}^{\frac{1}{2}} f (x) d x = - 4 a r c s i n (\frac{1}{4}) - \frac{\sqrt{15}}{4}$
Commented by maxmathsup by imad last updated on 06/May/19

$4) \int_{4}^{5} g (x) d x = \int_{4}^{5} \frac{2 + \sqrt{x - 3}}{2 - \sqrt{x - 3}} d x =_{\sqrt{x - 3} = t} \int_{1}^{\sqrt{2}} \frac{2 + t}{2 - t} (2 t) d t$ $= 2 \int_{1}^{\sqrt{2}} \frac{t^{2} + 2 t}{2 - t} d t = - 2 \int_{1}^{\sqrt{2}} \frac{t^{2} + 2 t}{t - 2} d t = - 2 \int_{1}^{\sqrt{2}} \frac{t^{2} - 4 + 4 + 2 t}{t - 2}$ $= - 2 \int_{1}^{\sqrt{2}} (t + 2 + \frac{2 t + 4}{t - 2}) d t = - 2 \int_{1}^{\sqrt{2}} (t + 2) d t - 4 \int_{1}^{\sqrt{2}} \frac{t + 2}{t - 2} d t b u t$ $\int_{1}^{\sqrt{2}} (t + 2) d t = {[\frac{t^{2}}{2} + 2 t]}_{1}^{\sqrt{2}} = 1 + 2 \sqrt{2} - \frac{1}{2} - 2 = 2 \sqrt{2} - \frac{3}{2}$ $\int_{1}^{\sqrt{2}} \frac{t + 2}{t - 2} d t = \int_{1}^{\sqrt{2}} \frac{t - 2 + 4}{t - 2} d t = \sqrt{2} - 1 + 4 {[l n ∣ t - 2 ∣]}_{1}^{\sqrt{2}} = \sqrt{2} - 1 + 4 (l n (2 - \sqrt{2}) \Rightarrow$ $\int_{4}^{5} g (x) d x = - 4 \sqrt{2} + 3 - 4 \sqrt{2} + 4 - 16 l n (2 - \sqrt{2}) = - 8 \sqrt{2} + 7 - 16 l n (2 - \sqrt{2}) .$
	Answered by Forkum Michael Choungong last updated on 05/May/19
	$for 1) 1) f(x)= x−(√(4−x^2 )) and g(x) =((2+(√(x−3)))/(2−(√(x−3)))) let f(x)=0 x−(√(4−x^2 ))=0 (√(4−x^2 ))=x 4−x^2 =x^2 4=2x^2 x=(√2) D_(f ) = { x:x ∈ R , x≠(√2)}$
	$f o r 1)$ $1) f (x) = x - \sqrt{4 - x^{2}} a n d g (x) = \frac{2 + \sqrt{x - 3}}{2 - \sqrt{x - 3}}$ $l e t f (x) = 0$ $x - \sqrt{4 - x^{2}} = 0$ $\sqrt{4 - x^{2}} = x$ $4 - x^{2} = x^{2}$ $4 = 2 x^{2}$ $x = \sqrt{2}$ $D_{f} = {x : x \in R, x \neq \sqrt{2}}$
	Commented by Mr X pcx last updated on 06/May/19

	$n o t c o r r e c t s i r . . .$
	Answered by MJS last updated on 06/May/19
	$D_f ={x∈R ∣ −2≤x≤2}; R_f ={y∈R ∣ −2(√2)≤y≤2} D_g ={x∈R ∣ x≥3∧x≠7}; R_g ={y∈R ∣ ∣y∣≥1} f(g(x)) is defined for −2≤g(x)≤2 ⇒ ⇒ D_(f(g(x))) ={x∈R ∣ 3≤x≤((31)/9) ∨ x≥39} g(f(x)) is defined for f(x)≥3∧f(x)≠7 ⇒ ⇒ D_(g(f(x))) ={}$
	$D_{f} = {x \in R ∣ - 2 ⩽ x ⩽ 2}; R_{f} = {y \in R ∣ - 2 \sqrt{2} ⩽ y ⩽ 2}$ $D_{g} = {x \in R ∣ x ⩾ 3 \land x \neq 7}; R_{g} = {y \in R ∣ ∣ y ∣⩾ 1}$ $f (g (x)) is defined for - 2 ⩽ g (x) ⩽ 2 \Rightarrow$ $\Rightarrow D_{f (g (x))} = {x \in R ∣ 3 ⩽ x ⩽ \frac{31}{9} \lor x ⩾ 39}$ $g (f (x)) is defined for f (x) ⩾ 3 \land f (x) \neq 7 \Rightarrow$ $\Rightarrow D_{g (f (x))} = {}$
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