Question-101778

Question Number 101778 by mathocean1 last updated on 04/Jul/20

Answered by mr W last updated on 04/Jul/20

f (x) = P (x)

\frac{x^{2} + 4 x + 1}{x + 1} = x^{2}

x^{3} - 4 x - 1 = 0

s a y g (x) = x^{3} - 4 x - 1

g (0) = - 1 < 0

\lim_{x \to + \infty} g (x) = + \infty > 0

\Rightarrow t h e r e i s a t l e a s t o n e v a l u e x_{A} \in (0, + \infty)

g (x_{A}) = 0

i . e . f (x) a n d P (x) i n t e r s e c t a t l e a s t

a t o n e p o i n t A (x_{A}, y_{A}) w i t h x_{A} > 0 a n d

y_{A} = x_{A}^{2} > 0 .

f o r r o o t s o f x^{3} - 4 x - 1 = 0 s e e Q 101658

Commented by mathocean1 last updated on 07/Jul/20

t h a n k y o u s i r .

Answered by 1549442205 last updated on 07/Jul/20

The cordinates of the intersection points

of two the curves are roots of the system :

{\begin{cases} y = \frac{x^{2} + 4 x + 1}{x + 1} \\ y = x^{2} \end{cases} \Leftrightarrow {\begin{cases} \frac{x^{2} + 4 x + 1}{x + 1} = x^{2} (1) \\ y = x^{2} (2) \end{cases}

(1) \Leftrightarrow x^{3} - 4 x - 1 = 0 \Leftrightarrow x = \frac{- 4 \sqrt{3}}{3} \cos \frac{\arccos \frac{- 3 \sqrt{3}}{16} + 2 k π}{3} with k \in {0, 1, 2}

(see to solve eqs . in the question 101658)

For k \in {0, 2} we get the negative roots

which rejected

For k = 1 we get x = 2.114907541

Thus, x_{A} = 2.1149, y_{A} = 4.4728

Commented by mathocean1 last updated on 06/Jul/20

T h a n k y o u s i r s!

P l e a s e i s t h e r e o t h e r w a y t o s o l v e

x^{3} - 4 x - 1 = 0 \dots i {d o n}^{'} t k n o w :

a r c c o s \dots i^{'} m s t i l l l e a r n e r i n t r i g o .

Commented by 1549442205 last updated on 07/Jul/20

\arccos or \cos^{- 1} is convert function

of \cos . Arccos m is a angle which

0 ⩽ Arccosm ⩽ π so that if \arccos m = φ

then \cos φ = m . Example : If cosx = \frac{1}{3}

then we have x = \arccos \frac{1}{3} . In other words,

Arccos a is the root of the equation

cosx = a . Sometimes ones write \cos^{- 1} m

instead of \arccos m

Commented by mathocean1 last updated on 07/Jul/20

t h a n k y o u v e r y m u c h s i r!