let-f-x-x-3-x-3-1-prove-that-f-have-one-root-real-0-and-0-1-2-2-factorize-f-x-inside-R-x-and-C-x-3-find-dx-f-x-

Question Number 104771 by mathmax by abdo last updated on 23/Jul/20

let f (x) = x^{3} + x - 3

1) prove that f have one root real α_{0} and α_{0} \in] 1, 2 [

2) factorize f (x) inside R [x] and C [x]

3) find \int \frac{dx}{f (x)}

Answered by abdomathmax last updated on 24/Jul/20

1) f continue and f^{^{'}} (x) = {3 x}^{2} + 1 > 0 \Rightarrow f is increazing

we have f (1) = - 1 and f (2) = 8 + 2 - 3 = 7 \Rightarrow

f (1) . f (2) < 0 \Rightarrow \exists α_{0} \in] 1, 2 [unique / f (α_{0}) = 0

2) f (x) = (x - α_{0}) (x^{2} + ax + b) \Rightarrow

x^{3} + {ax}^{2} + bx - α_{0} x^{2} - α_{o} a x - b α_{0} = x^{3} + x - 3 \Rightarrow

(a - α_{0}) x^{2} + (b - α_{o} a) x - b α_{0} = x - 3 \Rightarrow

{\begin{cases} a - α_{0} = 0 \\ b - α_{0} a = 1 and - b α_{0} = - 3 \Rightarrow \end{cases}

a = α_{0} and b = \frac{3}{α_{0}}

f (x) = (x - α_{0}) (x^{2} + α_{0} x + \frac{3}{α_{0}}) this is tbe factoruzation

at R [x]

x^{2} + α_{0} x + \frac{3}{α_{0}} = 0 \to Δ = α_{0}^{2} - \frac{12}{α_{0}} = \frac{α_{0}^{3} - 12}{α_{0}}

= \frac{- α_{o} + 3 - 12}{α_{0}} = - \frac{α_{0} + 9}{α_{0}} \Rightarrow z_{1} = - α_{0} + i \sqrt{\frac{α_{0} + 9}{α_{0}}}

z_{2} = - α_{0} - i \sqrt{\frac{α_{0} + 9}{α_{0}}}

and f (x) = (x - α_{0}) (x - z_{1}) (x - z_{2})

= (x - α_{0}) (x + α_{0} - i \sqrt{\frac{α_{0} + 9}{α_{0}}}) (x + α_{0} + i \sqrt{\frac{α_{0} + 9}{α_{0}}})