Given-f-x-x-4-ax-3-bx-2-cx-d-where-a-b-c-and-d-are-real-number-suppose-the-graph-f-x-intersects-the-graph-of-y-2x-1-at-x-1-2-3-Find-the-value-of-f-0-f-4-

Question Number 174806 by cortano1 last updated on 11/Aug/22

G i v e n f (x) = x^{4} + {a x}^{3} + {b x}^{2} + c x + d

w h e r e a, b, c a n d d a r e r e a l n u m b e r

s u p p o s e t h e g r a p h f (x) i n t e r s e c t s

t h e g r a p h o f y = 2 x - 1 a t x = 1, 2, 3 .

F i n d t h e v a l u e o f f (0) + f (4) .

Commented by Rasheed.Sindhi last updated on 11/Aug/22

You can't use 'macro parameter character #' in math mode

Answered by floor(10²Eta[1]) last updated on 11/Aug/22

\Rightarrow 1 + a + b + c + d = 1 \Rightarrow a + b + c + d = 0 (I)

16 + 8 a + 4 b + 2 c + d = 3 \Rightarrow 8 a + 4 b + 2 c + d = - 13 (II)

81 + 27 a + 9 b + 3 c + d = 5 (III)

\Rightarrow (II) - (I) = 7 a + 3 b + c = - 13 (IV)

(III) - (I) = 26 a + 8 b + 2 c = - 76 (V)

(V) - 2 (IV) = 12 a + 2 b = - 50 \Rightarrow 6 a + b = - 25 (VI)

f (0) + f (4) = 256 + 64 a + 16 b + 4 c + 2 d

= 256 + 48 a + 8 b + (16 a + 8 b + 4 c + 2 d) \to 2 (II)

= 256 + 48 a + 8 b - 26 = 230 + 8 (6 a + b) \to (VI)

= 230 - 8 (25) = 30