Given f(x)=(√(ax+b)) ; g(x)=px+q and f(x)<g(x) has solution 1<x<5. Give one example f(x) and g(x) satisfy the condition


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Question Number 130670 by EDWIN88 last updated on 27/Jan/21

$G i v e n f (x) = \sqrt{a x + b}; g (x) = p x + q$ $a n d f (x) < g (x) h a s s o l u t i o n$ $1 < x < 5 . G i v e o n e e x a m p l e f (x)$ $a n d g (x) s a t i s f y t h e c o n d i t i o n$
Commented byMJS_new last updated on 28/Jan/21

$do you have an example ? to me it seems$ $impossible with a, b, p, q, x \in R$
Commented byEDWIN88 last updated on 28/Jan/21

$y e s s i r . i t h i n k {i t}^{'} s p o s s i b l e$ $i f f (x) > g (x)$
	Answered by MJS_new last updated on 28/Jan/21
	$with f(x)>g(x) we start at the borders x=1 ⇒ (√(a+b))=p+q x=5 ⇒ (√(5a+b))=5p+q ⇒ p=(((√(5a+b))−(√(a+b)))/4)∧q=((5(√(a+b))−(√(5a+b)))/4) ⇒ (√(ax+b))>(((√(5a+b))−(√(a+b)))/4)x+((5(√(a+b))−(√(5a+b)))/4); 1<x<5 test this for x=3 (√(3a+b))>(((√(5a+b))+(√(a+b)))/2) 2(√(3a+b))>(√(5a+b))+(√(a+b)) t^2 =3a+b∧t>0 ⇒ b=t^2 −3a 2t>(√(t^2 +2a))+(√(t^2 −2a)) 4t^2 >2t^2 +2(√(t^2 +2a))(√(t^2 −2a)) t^2 >(√(t^2 +2a))(√(t^2 −2a)) t^4 >t^4 −4a^4 true for a≠0 ⇒ we can choose a, b with ax+b≥0∧1<x<5 ⇔ ⇔ b≥−ax∧1<x<5 ⇒ { ((b>−5a; a<0)),((b>−a; a>0)) :} examples (1) a=−3∧b=16 ⇒ p=((1−(√(13)))/4)∧q=((−1+5(√(13)))/4) (√(−3x+16))>((1−(√(13)))/4)x−((1−5(√(13)))/4); 1<x<5 (2) a=5∧b=−1 ⇒ p=((−1+(√6))/2)∧q=((5−(√6))/2) (√(5x−1))>−((1−(√6))/2)x+((5−(√6))/2); 1<x<5 plot them to see it′s true$
	$with f (x) > g (x) we start at the borders$ $x = 1 \Rightarrow \sqrt{a + b} = p + q$ $x = 5 \Rightarrow \sqrt{5 a + b} = 5 p + q$ $\Rightarrow$ $p = \frac{\sqrt{5 a + b} - \sqrt{a + b}}{4} \land q = \frac{5 \sqrt{a + b} - \sqrt{5 a + b}}{4}$ $\Rightarrow$ $\sqrt{a x + b} > \frac{\sqrt{5 a + b} - \sqrt{a + b}}{4} x + \frac{5 \sqrt{a + b} - \sqrt{5 a + b}}{4}; 1 < x < 5$ $test this for x = 3$ $\sqrt{3 a + b} > \frac{\sqrt{5 a + b} + \sqrt{a + b}}{2}$ $2 \sqrt{3 a + b} > \sqrt{5 a + b} + \sqrt{a + b}$ $t^{2} = 3 a + b \land t > 0 \Rightarrow b = t^{2} - 3 a$ $2 t > \sqrt{t^{2} + 2 a} + \sqrt{t^{2} - 2 a}$ $4 t^{2} > 2 t^{2} + 2 \sqrt{t^{2} + 2 a} \sqrt{t^{2} - 2 a}$ $t^{2} > \sqrt{t^{2} + 2 a} \sqrt{t^{2} - 2 a}$ $t^{4} > t^{4} - 4 a^{4}$ $true for a \neq 0$ $\Rightarrow$ $we can choose a, b with a x + b ⩾ 0 \land 1 < x < 5 \Leftrightarrow$ $\Leftrightarrow b ⩾ - a x \land 1 < x < 5 \Rightarrow {\begin{cases} b > - 5 a; a < 0 \\ b > - a; a > 0 \end{cases}$ $examples$ $(1) a = - 3 \land b = 16 \Rightarrow p = \frac{1 - \sqrt{13}}{4} \land q = \frac{- 1 + 5 \sqrt{13}}{4}$ $\sqrt{- 3 x + 16} > \frac{1 - \sqrt{13}}{4} x - \frac{1 - 5 \sqrt{13}}{4}; 1 < x < 5$ $(2) a = 5 \land b = - 1 \Rightarrow p = \frac{- 1 + \sqrt{6}}{2} \land q = \frac{5 - \sqrt{6}}{2}$ $\sqrt{5 x - 1} > - \frac{1 - \sqrt{6}}{2} x + \frac{5 - \sqrt{6}}{2}; 1 < x < 5$ $plot them to see {it}^{'} s true$
	Commented byliberty last updated on 28/Jan/21

	Commented byliberty last updated on 28/Jan/21

	$correct sir$
	Commented byEDWIN88 last updated on 28/Jan/21

	$i g o t \sqrt{- 4 x + 20} > - x + 5 s i r$
	Commented byEDWIN88 last updated on 28/Jan/21

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