Question-107461

Question Number 107461 by Don08q last updated on 10/Aug/20

Commented by john santu last updated on 11/Aug/20

(i i i) y = x^{3} + (k + 2) x^{2} - 3 k x + 5

\Rightarrow y^{'} = 3 x^{2} + 2 (k + 2) x - 3 k

\Rightarrow Δ = 4 {(k + 2)}^{2} - 4.3 . (- 3 k) = 0

\Rightarrow {(k + 2)}^{2} + 9 k = 0

\Rightarrow k^{2} + 13 k + 4 = 0

k = \frac{- 13 \pm \sqrt{169 - 4.1.4}}{2} = \frac{- 13 \pm \sqrt{153}}{2}

k = \frac{- 13 \pm 3 \sqrt{17}}{2} .

Answered by john santu last updated on 11/Aug/20

⋇ JS ⋇

\Rightarrow y = {a x}^{3} + {b x}^{2} + c x + d \Rightarrow h a s a

v e r t e x i f y^{'} = 0

\Rightarrow y^{'} = 3 {a x}^{2} + 2 b x + c = 0

\Rightarrow x = \frac{- 2 b \pm \sqrt{4 b^{2} - 4.3 a . c}}{6 a}

x = \frac{- 2 b \pm 2 \sqrt{b^{2} - 3 a c}}{6 a} = \frac{- b \pm \sqrt{b^{2} - 3 a c}}{3 a}

Commented by john santu last updated on 11/Aug/20

(i i) n o v e r t e x i f b^{2} - 3 a c < 0

\Rightarrow (b - \sqrt{3 a c}) (b + \sqrt{3 a c}) < 0

\Rightarrow - \sqrt{3 a c} < b < \sqrt{3 a c}

Commented by Don08q last updated on 11/Aug/20

T h a n k y o u S i r

Answered by 1549442205PVT last updated on 11/Aug/20

We need to have the hypothesis that

a, b \neq 0 . Then denote by C the graph of

y = {ax}^{3} + {bx}^{2} + cx + d . We have

y^{'} = {3 ax}^{2} + 2 bx + c = 0 \Leftrightarrow x_{1, 2} = \frac{- b \pm \sqrt{b^{2} - 3 ac}}{3 a}

y^{″} = 6 ax + 2 b = 0 \Leftrightarrow x_{0} = - b / 3 a

i) if Δ = b^{2} - 3 ac > 0 the eqn . y^{'} = 0 has

two real roots and y^{'} change the sign when

passes through its roots, so the curve

C has two extremums points (vertexes)

ii) If b^{2} - 3 ac = 0 then x_{0} = x_{1} = x_{2} = - b / 3 a .

Then the given function has no extremum

(vertex) because y^{'} {don}^{'} t change the

sign when go through x_{0} . Hence, the

condition need and enough for the curve

of degree 3 has a saddle point be

Δ = b^{2} - 3 ac = 0

iii) The curve y = x^{3} + (k + 2) x^{2} - 3 kx + 5 (C_{k})

has the saddle point if and only if

y^{'} = {3 x}^{2} + 2 (k + 2) x - 3 k = 0 has unique root

\Leftrightarrow Δ^{'} = {(k + 2)}^{2} + 9 k = 0 \Leftrightarrow k^{2} + 13 k + 4 = 0

\Leftrightarrow k_{1, 2} = \frac{- 13 \pm 3 \sqrt{17}}{2} . Then x_{0} = \frac{- (k + 2)}{3}

+) for k_{1} = \frac{- 13 + 3 \sqrt{17}}{2} \Rightarrow x_{0} = - \frac{- 9 + 3 \sqrt{17}}{6}

+) For k_{2} = \frac{- 13 - 3 \sqrt{17}}{2} \Rightarrow x_{0} = \frac{9 + 3 \sqrt{17}}{6}

Thus, C_{k} has the saddle point if and

only if k = \frac{- 13 \pm 3 \sqrt{17}}{2}

Commented by Don08q last updated on 11/Aug/20

C o m p r e h e n s i v e e x p l a n a t i o n . T h a n k y o u

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