f(x)= (√( x −a)) + (√(3a −x)) with ( a>0) is given .If , f_( max) . f


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Question Number 186265 by mnjuly1970 last updated on 02/Feb/23

$f (x) = \sqrt{x - a} + \sqrt{3 a - x} w i t h (a > 0)$ $i s g i v e n . I f, f_{m a x} . f_{m i n} = \sqrt{32}$ $f i n d, ε a ε = ?$
	Answered by mahdipoor last updated on 02/Feb/23
	$(df/dx)=(1/(2(√(x−a))))+((−1)/(2(√(3a−x))))=0⇒(√(3a−x))=(√(x−a)) ⇒x=2a D_f =[a,3a] f(a)=f(3a)=(√(2a)) f(2a)=2(√a) f_(min) .f_(max) =(√(8a^2 ))=(√(32)) ⇒a=2 way 2: f^2 =2a+2(√(4ax−x^2 −3a^2 )) g(x)= (√(4ax−x^2 −3a^2 )) { ((min=0)),((max=g(((−4a)/(−1×2)))=a)) :} min f^2 ×max f^2 =2a×4a=((√(32)))^2 ⇒a=2$
	$\frac{d f}{d x} = \frac{1}{2 \sqrt{x - a}} + \frac{- 1}{2 \sqrt{3 a - x}} = 0 \Rightarrow \sqrt{3 a - x} = \sqrt{x - a}$ $\Rightarrow x = 2 a$ $D_{f} = [a, 3 a]$ $f (a) = f (3 a) = \sqrt{2 a} f (2 a) = 2 \sqrt{a}$ $f_{m i n} . f_{m a x} = \sqrt{8 a^{2}} = \sqrt{32} \Rightarrow a = 2$ $w a y 2 :$ $f^{2} = 2 a + 2 \sqrt{4 a x - x^{2} - 3 a^{2}}$ $g (x) = \sqrt{4 a x - x^{2} - 3 a^{2}} {\begin{cases} m i n = 0 \\ m a x = g (\frac{- 4 a}{- 1 \times 2}) = a \end{cases}$ $m i n f^{2} \times m a x f^{2} = 2 a \times 4 a = {(\sqrt{32})}^{2} \Rightarrow a = 2$
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