f(x)=x^(2014) +2x^(2013) +3x^(2012) +4x^(2011) +...+2014x+2015 min f(x)=?


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Question Number 157952 by cortano last updated on 30/Oct/21

$f (x) = x^{2014} + 2 x^{2013} + 3 x^{2012} + 4 x^{2011} + . . . + 2014 x + 2015$ $m i n f (x) = ?$
	Answered by aleks041103 last updated on 31/Oct/21

	$f (x) = \underset{i = 1}{\sum^{2015}} {i x}^{2015 - i} = - x^{2016} \underset{i = 1}{\sum^{2015}} (- i) x^{- i - 1} =$ $= - x^{2016} \frac{d}{d x} (\underset{i = 1}{\sum^{2015}} x^{- i})$ $\underset{i = 1}{\sum^{2015}} x^{- i} = \frac{1}{x} \underset{i = 0}{\sum^{2014}} {(\frac{1}{x})}^{i} = \frac{1}{x} \frac{{(\frac{1}{x})}^{2015} - 1}{\frac{1}{x} - 1} =$ $= - \frac{x^{- 2015} - 1}{x - 1}$ $\Rightarrow f (x) = - x^{2016} \frac{d}{d x} (\frac{1 - x^{- 2015}}{x - 1}) =$ $= - x^{2016} \frac{2015 x^{- 2016} (x - 1) - 1 + x^{- 2015}}{{(x - 1)}^{2}} =$ $= - \frac{2015 (x - 1) - x^{2016} + x}{{(x - 1)}^{2}} = \frac{x^{2016} - 2016 x + 2015}{{(x - 1)}^{2}}$ $\Rightarrow f (x) = \frac{x^{2016} - 2016 x + 2015}{{(x - 1)}^{2}}$ $f^{'} (x) = \frac{(2016 x^{2015} - 2016) {(x - 1)}^{2} - 2 (x - 1) (x^{2016} - 2016 x + 2015)}{{(x - 1)}^{4}} =$ $= \frac{2016 (x^{2015} - 1) (x - 1) - 2 x^{2016} + 2.2016 x - 2.2015}{{(x - 1)}^{3}} =$ $= \frac{2016 (x^{2016} - x^{2015} - x + 1) - 2 x^{2016} + 2.2016 x - 2.2015}{{(x - 1)}^{3}} =$ $= \frac{2014 x^{2016} - 2016 x^{2015} + 2016 x - 2014}{{(x - 1)}^{3}}$ $F o r e x t r e m u m w e n e e d f^{'} = 0 \Rightarrow$ $2014 x^{2016} - 2016 x^{2015} + 2016 x - 2014 = 0$ $T h i s p o l y n o m e h a s o n l y 2 r e a l s o l u t i o n s :$ $x = + 1; - 1 .$ $f^{'} (1) = \underset{x \to 1}{l i m} \frac{2014 x^{2016} - 2016 x^{2015} + 2016 x - 2014}{{(x - 1)}^{3}} =$ $L^{'} H = \underset{x \to 1}{l i m} \frac{2016 (2014 x^{2015} - 2015 x^{2014} + 1)}{3 {(x - 1)}^{2}} =$ $L^{'} H = \frac{2016}{3} \underset{x \to 1}{l i m} \frac{2015.2014 x^{2013} (x - 1)}{2 (x - 1)} =$ $= \frac{2016.2015.2014}{6} \neq 0$ $\Rightarrow T h i s l e a v e s f o r p o s s i b l e e x t r e m a o n l y x = - 1$ $I t c a n b e e a s i l y s e e n t h a t :$ $\underset{x \to \pm \infty}{l i m} f (x) \to + \infty$ $T h e n f (x) c a n o n l y h a v e a m i n i m u m$ $a t x = - 1 .$ $T h e n t h e a n s w e r w i l l b e :$ $m i n (f) = f (- 1) = \frac{{(- 1)}^{2016} - 2016 (- 1) + 2015}{{(- 1 - 1)}^{2}} =$ $= \frac{1 + 2016 + 2015}{4} = 1008$ $\Rightarrow m i n (f) = 1008$
	Commented by aleks041103 last updated on 31/Oct/21
	$Additional: Solutions to the polynomial 2014x^(2016) −2016x^(2015) +2016x−2014=0 p(x)=2014x^(2016) −2016x^(2015) +2016x−2014 Solution: Let′s find the extremums p′=2016.2014x^(2015) −2016.2015x^(2014) +2016= =2016(2014x^(2015) −2015x^(2014) +1)=2016g(x) let′s analyze g(x): g′(x)=2014.2015x^(2013) (x−1) g(x) can have extrema only at x=0 and x=1. g′′(x)=2014.2015x^(2012) (2014x−2013) g′′(1)=2014.2015>0 ⇒g(1)=0 is a min of g(x) g^((n)) (x)=2014.2015D^((n−1)) (x^(2014) −x^(2013) )= =2014.2015(((2014!)/((2015−n)!))x^(2015−n) −((2013!)/((2014−n)!))x^(2014−n) ) ⇒g^((n)) (0)= { ((0, n=1,2,...,2013,2016,...)),((−2015!, n=2014)),((2014.2015!, n=2015)) :} Since the smallest n for which g^((n)) (0)≠0 is even, g(0)=1 is a max of g(x). By the graph of g(x), which we vaugely construct using the analysis we did on g(x), we see that there are 2 points where g(x)=0 − at x=1 and at some x=x_1 <0. Then p′(x=1;x_1 )=0. p′′(x)=2016.2015.2014(x^(2014) −x^(2013) ) Then at x_1 <0 we see that p′′(x)>0. Therefore p(x) has a minimum at x=x_1 . p′′(x=1)=0 p′′′(x)=2016.2015.2014x^(2012) (2014x−2013) ⇒p′′′(x=1)=2016.2015.2014>0 ⇒p(x) has a saddle point at x=1. Then we construct vaugely the graph of p(x) and we see, that p(x)=0 has only 2 solutions. They can be found by inspection: p(1)=p(−1)=0$
	$A d d i t i o n a l :$ $S o l u t i o n s t o t h e p o l y n o m i a l$ $2014 x^{2016} - 2016 x^{2015} + 2016 x - 2014 = 0$ $p (x) = 2014 x^{2016} - 2016 x^{2015} + 2016 x - 2014$ $S o l u t i o n :$ ${L e t}^{'} s f i n d t h e e x t r e m u m s$ $p^{'} = 2016.2014 x^{2015} - 2016.2015 x^{2014} + 2016 =$ $= 2016 (2014 x^{2015} - 2015 x^{2014} + 1) = 2016 g (x)$ ${l e t}^{'} s a n a l y z e g (x) :$ $g^{'} (x) = 2014.2015 x^{2013} (x - 1)$ $g (x) c a n h a v e e x t r e m a o n l y a t x = 0 a n d$ $x = 1 .$ $g^{″} (x) = 2014.2015 x^{2012} (2014 x - 2013)$ $g^{″} (1) = 2014.2015 > 0$ $\Rightarrow g (1) = 0 i s a m i n o f g (x)$ $g^{(n)} (x) = 2014.2015 D^{(n - 1)} (x^{2014} - x^{2013}) =$ $= 2014.2015 (\frac{2014!}{(2015 - n)!} x^{2015 - n} - \frac{2013!}{(2014 - n)!} x^{2014 - n})$ $\Rightarrow g^{(n)} (0) = {\begin{cases} 0, n = 1, 2, . . ., 2013, 2016, . . . \\ - 2015!, n = 2014 \\ 2014.2015!, n = 2015 \end{cases}$ $S i n c e t h e s m a l l e s t n f o r w h i c h g^{(n)} (0) \neq 0$ $i s e v e n, g (0) = 1 i s a m a x o f g (x) .$ $B y t h e g r a p h o f g (x), w h i c h w e v a u g e l y c o n s t r u c t$ $u s i n g t h e a n a l y s i s w e d i d o n g (x), w e s e e t h a t t h e r e a r e$ $2 p o i n t s w h e r e g (x) = 0 - a t x = 1 a n d a t s o m e x = x_{1} < 0 .$ $T h e n p^{'} (x = 1; x_{1}) = 0 .$ $p^{″} (x) = 2016.2015.2014 (x^{2014} - x^{2013})$ $T h e n a t x_{1} < 0 w e s e e t h a t p^{″} (x) > 0 .$ $T h e r e f o r e p (x) h a s a m i n i m u m a t x = x_{1} .$ $p^{″} (x = 1) = 0$ $p^{‴} (x) = 2016.2015.2014 x^{2012} (2014 x - 2013)$ $\Rightarrow p^{‴} (x = 1) = 2016.2015.2014 > 0$ $\Rightarrow p (x) h a s a s a d d l e p o i n t a t x = 1 .$ $T h e n w e c o n s t r u c t v a u g e l y t h e g r a p h o f$ $p (x) a n d w e s e e, t h a t p (x) = 0 h a s o n l y$ $2 s o l u t i o n s . T h e y c a n b e f o u n d b y i n s p e c t i o n :$ $p (1) = p (- 1) = 0$
	Commented by aleks041103 last updated on 31/Oct/21

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