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.


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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

	$yes . . . thanks . . .$
	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

	$y e s, t h a n k s!$
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