f-x-2-f-x-1-2x-2-14-f-x-

Question Number 118740 by Jamshidbek2311 last updated on 19/Oct/20

f (x + 2) + f (x - 1) = 2 x^{2} + 14

f (x) = ?

Commented by PRITHWISH SEN 2 last updated on 19/Oct/20

x = x + 1

f (x + 3) + f (x) = {2 x}^{2} + 4 x + 16 \dots . . (i)

x = x - 2

f (x) + f (x - 3) = {2 x}^{2} - 8 x + 22 \dots (ii)

adding (i) & (ii)

f (x + 3) + f (x - 3) + 2 f (x) = 4 f (x) + 2 \times 3^{2} = {4 x}^{2} - 4 x + 38

\Rightarrow f (x) = \frac{{4 x}^{2} - 4 x + 38 - 2 \times 3^{2}}{4} = x^{2} - x + 5

Answered by bemath last updated on 19/Oct/20

l e t f (x) = {p x}^{2} + q x + r

f (x + 2) = {p x}^{2} + 4 p x + 4 p + q x + 2 q + r

= {p x}^{2} + (4 p + q) x + 4 p + 2 q + r

f (x - 1) = {p x}^{2} - 2 p x + p + q x - q + r

= {p x}^{2} + (- 2 p + q) x + p - q + r

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

2 {p x}^{2} + (2 p + 2 q) x + 5 p + q + 2 r = 2 x^{2} + 14

{\begin{cases} 2 p = 2 \Rightarrow p = 1 \\ 2 p + 2 q = 0; q = - 1 \\ 5 p + q + 2 r = 14; 4 + 2 r = 14; r = 5 \end{cases}

∴ f (x) = x^{2} - x + 5

Commented by Jamshidbek2311 last updated on 19/Oct/20

w r o n g .

Commented by benjo_mathlover last updated on 19/Oct/20

c h e c k i n g

f (x) = x^{2} - x + 5 \to {\begin{cases} f (x + 2) = x^{2} + 4 x + 4 - x - 2 + 5 = x^{2} + 3 x + 7 \\ f (x - 1) = x^{2} - 2 x + 1 - x + 1 + 5 = x^{2} - 3 x + 7 \end{cases}

t h e n f (x + 2) + f (x - 1) = 2 x^{2} + 14

C o r r e c t .