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

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 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 \in R

Commented by EDWIN88 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 \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 by liberty last updated on 28/Jan/21

Commented by liberty last updated on 28/Jan/21

correct sir

Commented by EDWIN88 last updated on 28/Jan/21

i g o t \sqrt{- 4 x + 20} > - x + 5 s i r

Commented by EDWIN88 last updated on 28/Jan/21

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