Find-the-polynomial-P-x-of-least-degree-that-has-a-maximum-equal-to-6-at-x-1-and-minimum-equal-to-2-at-x-3-

Question Number 122349 by liberty last updated on 16/Nov/20

Find the polynomial P (x) of least degree

that has a maximum equal to 6 at x = 1

and minimum equal to 2 at x = 3 .

Commented by benjo_mathlover last updated on 17/Nov/20

m y a n s w e r :

t h e p o l y n o m i a l o f t h e l e a s t d e g r e e

w i t h r o o t s x_{1} = 1 a n d x_{2} = 3 h a s t h e

f o r m a (x - 1) (x - 3) . H e n c e

P^{'} (x) = a (x - 1) (x - 3) = a (x^{2} - 4 x + 3)

S i n c e a t t h e p o i n t x = 1 t h e r e m u s t b e

P (1) = 6, w e h a v e

P (x) = \underset{1}{\int^{x}} P^{'} (x) d x + 6

P (x) = a \underset{1}{\int^{x}} (x^{2} - 4 x + 3) d x + 6

P (x) = a (\frac{x^{3}}{3} - 2 x^{2} + 3 x - \frac{4}{3}) + 6 .

T h e c o e f f i c i e n t a i s d e t e r m i n e d

f r o m t h e c o n d i t i o n P (3) = 2, w h e n c e

a = 3, H e n c e P (x) = x^{3} - 6 x^{2} + 9 x + 2 .

Answered by mr W last updated on 17/Nov/20

f^{'} (x) = k (x - 1) (x - 3) = k (x^{2} - 4 x + 3)

f (x) = k (\frac{x^{3}}{3} - 2 x^{2} + 3 x + c)

f (1) = k (\frac{4}{3} + c) = 6

f (3) = k c = 2

\Rightarrow k = 3

\Rightarrow c = \frac{2}{3}

\Rightarrow f (x) = x^{3} - 6 x^{2} + 9 x + 2

Commented by mr W last updated on 16/Nov/20

Commented by liberty last updated on 16/Nov/20

yes \dots thanks \dots

Commented by benjo_mathlover last updated on 17/Nov/20

t y p o s i r . w e g e t k = 3 . i t s h o u l d

b e f (x) = 3 (\frac{x^{3}}{3} - 2 x^{2} + 3 x + \frac{2}{3})

f (x) = x^{3} - 6 x^{3} + 9 x + 2

Commented by mr W last updated on 17/Nov/20

y e s, t h a n k s!

Facebook Tweet Pin