Given-f-x-ax-b-g-x-px-q-and-f-x-lt-g-x-has-solution-1-lt-x-lt-5-Give-one-example-f-x-and-g-x-satisfy-the-condition-

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

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

$$\:{Given}\:{f}\left({x}\right)=\sqrt{{ax}+{b}}\:;\:{g}\left({x}\right)={px}+{q} \\ $$$${and}\:{f}\left({x}\right)<{g}\left({x}\right)\:{has}\:{solution} \\ $$$$\mathrm{1}<{x}<\mathrm{5}.\:{Give}\:{one}\:{example}\:{f}\left({x}\right) \\ $$$${and}\:{g}\left({x}\right)\:{satisfy}\:{the}\:{condition} \\ $$

Commented by MJS_new last updated on 28/Jan/21

do you have an example? to me it seems impossible with a, b, p, q, x ∈R

$$\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{an}\:\mathrm{example}?\:\mathrm{to}\:\mathrm{me}\:\mathrm{it}\:\mathrm{seems} \\ $$$$\mathrm{impossible}\:\mathrm{with}\:{a},\:{b},\:{p},\:{q},\:{x}\:\in\mathbb{R} \\ $$

Commented by EDWIN88 last updated on 28/Jan/21

yes sir. i think it′s possible if f(x)>g(x)

$${yes}\:{sir}.\:{i}\:{think}\:{it}'{s}\:{possible} \\ $$$${if}\:{f}\left({x}\right)>{g}\left({x}\right) \\ $$

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

$$\mathrm{with}\:{f}\left({x}\right)>{g}\left({x}\right)\:\mathrm{we}\:\mathrm{start}\:\mathrm{at}\:\mathrm{the}\:\mathrm{borders} \\ $$$${x}=\mathrm{1}\:\Rightarrow\:\sqrt{{a}+{b}}={p}+{q} \\ $$$${x}=\mathrm{5}\:\Rightarrow\:\sqrt{\mathrm{5}{a}+{b}}=\mathrm{5}{p}+{q} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\sqrt{\mathrm{5}{a}+{b}}−\sqrt{{a}+{b}}}{\mathrm{4}}\wedge{q}=\frac{\mathrm{5}\sqrt{{a}+{b}}−\sqrt{\mathrm{5}{a}+{b}}}{\mathrm{4}} \\ $$$$\Rightarrow \\ $$$$\sqrt{{ax}+{b}}>\frac{\sqrt{\mathrm{5}{a}+{b}}−\sqrt{{a}+{b}}}{\mathrm{4}}{x}+\frac{\mathrm{5}\sqrt{{a}+{b}}−\sqrt{\mathrm{5}{a}+{b}}}{\mathrm{4}};\:\mathrm{1}<{x}<\mathrm{5} \\ $$$$\mathrm{test}\:\mathrm{this}\:\mathrm{for}\:{x}=\mathrm{3} \\ $$$$\sqrt{\mathrm{3}{a}+{b}}>\frac{\sqrt{\mathrm{5}{a}+{b}}+\sqrt{{a}+{b}}}{\mathrm{2}} \\ $$$$\mathrm{2}\sqrt{\mathrm{3}{a}+{b}}>\sqrt{\mathrm{5}{a}+{b}}+\sqrt{{a}+{b}} \\ $$$${t}^{\mathrm{2}} =\mathrm{3}{a}+{b}\wedge{t}>\mathrm{0}\:\Rightarrow\:{b}={t}^{\mathrm{2}} −\mathrm{3}{a} \\ $$$$\mathrm{2}{t}>\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{a}}+\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{a}} \\ $$$$\mathrm{4}{t}^{\mathrm{2}} >\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{a}}\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{a}} \\ $$$${t}^{\mathrm{2}} >\sqrt{{t}^{\mathrm{2}} +\mathrm{2}{a}}\sqrt{{t}^{\mathrm{2}} −\mathrm{2}{a}} \\ $$$${t}^{\mathrm{4}} >{t}^{\mathrm{4}} −\mathrm{4}{a}^{\mathrm{4}} \\ $$$$\mathrm{true}\:\mathrm{for}\:{a}\neq\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{choose}\:{a},\:{b}\:\mathrm{with}\:{ax}+{b}\geqslant\mathrm{0}\wedge\mathrm{1}<{x}<\mathrm{5}\:\Leftrightarrow \\ $$$$\Leftrightarrow\:{b}\geqslant−{ax}\wedge\mathrm{1}<{x}<\mathrm{5}\:\Rightarrow\:\begin{cases}{{b}>−\mathrm{5}{a};\:{a}<\mathrm{0}}\\{{b}>−{a};\:{a}>\mathrm{0}}\end{cases} \\ $$$$\mathrm{examples} \\ $$$$\left(\mathrm{1}\right)\:{a}=−\mathrm{3}\wedge{b}=\mathrm{16}\:\Rightarrow\:{p}=\frac{\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{4}}\wedge{q}=\frac{−\mathrm{1}+\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{4}} \\ $$$$\sqrt{−\mathrm{3}{x}+\mathrm{16}}>\frac{\mathrm{1}−\sqrt{\mathrm{13}}}{\mathrm{4}}{x}−\frac{\mathrm{1}−\mathrm{5}\sqrt{\mathrm{13}}}{\mathrm{4}};\:\mathrm{1}<{x}<\mathrm{5} \\ $$$$\left(\mathrm{2}\right)\:{a}=\mathrm{5}\wedge{b}=−\mathrm{1}\:\Rightarrow\:{p}=\frac{−\mathrm{1}+\sqrt{\mathrm{6}}}{\mathrm{2}}\wedge{q}=\frac{\mathrm{5}−\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{5}{x}−\mathrm{1}}>−\frac{\mathrm{1}−\sqrt{\mathrm{6}}}{\mathrm{2}}{x}+\frac{\mathrm{5}−\sqrt{\mathrm{6}}}{\mathrm{2}};\:\mathrm{1}<{x}<\mathrm{5} \\ $$$$\mathrm{plot}\:\mathrm{them}\:\mathrm{to}\:\mathrm{see}\:\mathrm{it}'\mathrm{s}\:\mathrm{true} \\ $$

Commented by liberty last updated on 28/Jan/21