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) }](https://www.tinkutara.com/question/Q212140.png)
$$ \\ $$$$\:\mathrm{I}{f},\:\:\:\:\:{f}\left({x}\right)=−\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:−\mathrm{3}\: \\ $$$$\:\:\:\:\: \\ $$$$,\:{g}\left({x}\right)=\:\begin{cases}{\:\sqrt{\mathrm{7}−{x}}\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\:<\mathrm{7}}\\{\:\:\lfloor\:\mathrm{5}{x}\:\rfloor\:−\mathrm{5}{x}\:\:\:\:\:\:\:\:{x}\geqslant\mathrm{7}}\end{cases}\:\:\: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:{R}_{{fog}} \:=\:\left({a}\:,{b}\right]\:\: \\ $$$$\:\:\:\:\:\:\:{find}\:\:{the}\:{value}\:{of}\:\:\:{b}−{a} \\ $$$$\:\:\:\:\:\:{R}_{{fog}} \:=\:\left\{\:\left({fog}\right)\left({x}\right)\mid\:{x}\in\:{D}_{{fog}} \:\right\} \\ $$
Answered by mehdee7396 last updated on 04/Oct/24
![R_g =(−1,0]∪(0,(√7)]=(−1,(√7)] f(x)=−(x−2)^2 +1 x∈D_g ⇒−1<x≤(√7)⇒−3<x−2≤−2+(√7) 0≤(x−2)^2 <9 −9<−(x−2)^2 ≤0 −8<R_(fog) ≤1 ⇒b−a=9 ✓](https://www.tinkutara.com/question/Q212161.png)
$${R}_{{g}} =\left(−\mathrm{1},\mathrm{0}\right]\cup\left(\mathrm{0},\sqrt{\mathrm{7}}\right]=\left(−\mathrm{1},\sqrt{\mathrm{7}}\right] \\ $$$${f}\left({x}\right)=−\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{1} \\ $$$${x}\in{D}_{{g}} \Rightarrow−\mathrm{1}<{x}\leqslant\sqrt{\mathrm{7}}\Rightarrow−\mathrm{3}<{x}−\mathrm{2}\leqslant−\mathrm{2}+\sqrt{\mathrm{7}} \\ $$$$\mathrm{0}\leqslant\left({x}−\mathrm{2}\right)^{\mathrm{2}} <\mathrm{9} \\ $$$$−\mathrm{9}<−\left({x}−\mathrm{2}\right)^{\mathrm{2}} \leqslant\mathrm{0} \\ $$$$−\mathrm{8}<{R}_{{fog}} \leqslant\mathrm{1} \\ $$$$\Rightarrow{b}−{a}=\mathrm{9}\:\checkmark \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 04/Oct/24
$$\:\:\:\:{thabks}\:{alot}\:{sir}\:{mehdee} \\ $$