let-f-x-be-a-dolvnomial-of-degree-4-such-that-f-1-1-f-2-2-f-3-3-f-4-4-then-f-6-

Question Number 98842 by M±th+et+s last updated on 16/Jun/20

l e t f (x) b e a d o l v n o m i a l o f d e g r e e 4

s u c h t h a t f (1) = 1, f (2) = 2, f (3) = 3, f (4) = 4

t h e n f (6) = ?

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

s e e a l s o Q 98268

Answered by maths mind last updated on 16/Jun/20

l e t p (x) = f (x) - x \Rightarrow P (x) \in R_{4} [X]

\Rightarrow p (1) = p (2) = p (3) = p (4) = 0

1, 2, 3, 4 a r e r o o t s o f p

p (x) = a (x - 1) (x - 2) (x - 3) (x - 4)

f (x) = a (x - 1) (x - 2) (x - 3) (x - 4) + x

f (6) = 6 + 120 a

Commented by M±th+et+s last updated on 16/Jun/20

t h a n k y o u s i r

Answered by Rio Michael last updated on 17/Jun/20

Following the above procedure,

let p (x) = f (x) - x

\Rightarrow p (1) = f (1) - 1 = 0

p (2) = f (2) - 2 = 0

p (3) = f (3) - 3 = 0

p (4) = f (4) - 4 = 0

\Rightarrow f (x) = p (x) + x where p (x) = (x - 1) (x - 2) (x - 3) (x - 4)

\Rightarrow f (x) = (x - 1) (x - 2) (x - 3) (x - 4) + x

f (6) = (5) (4) (3) (2) + 6 = 126

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

4 c o n d i t i o n s a r e n o t e n o u g h t o

u i n q u e l y d e f i n e a 4 t h d e g r e e

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

u n i q u e s o l u t i o n f o r f (x) . w e c a n o n l y

s a y

f (x) = k (x - 1) (x - 2) (x - 3) (x - 4) + x

w i t h k = c o n s t a n t \neq 0

f (6) = 120 k + 6

Commented by Rio Michael last updated on 17/Jun/20

thanks sir, i got it now \dots