Given-f-R-R-is-a-quadratic-polynomial-f-1-1-f-2-1-2-and-f-3-1-3-Find-f-4-

Question Number 200167 by hardmath last updated on 15/Nov/23

Given f : R \to R is a quadratic polynomial

f (1) = 1, f (2) = \frac{1}{2} and f (3) = \frac{1}{3}

Find : f (4) = ?

Answered by witcher3 last updated on 15/Nov/23

\Leftrightarrow f (1) - 1 = 2 f (2) - 1 = 3 f (3) - 1 = 0

let xf (x) - 1 = g (x) \Rightarrow \deg (g) = 3

g (1) = g (2) = g (3) = 0

\Rightarrow g (x) = a (x - 1) (x - 2) (x - 3)

xf (x) = 1 + g (x)

\Rightarrow 1 + g (0) = 0 \Rightarrow - 6 a + 1 = 0

a = \frac{1}{6}

f (x) = \frac{1}{x} (\frac{1}{6} (x - 3) (x - 2) (x - 1) + 1)

= g^{'} (x) = \frac{x^{2}}{6} - x + \frac{11}{6} = f (x)

Answered by Rasheed.Sindhi last updated on 15/Nov/23

l e t f (x) = {a x}^{2} + b x + c

f (1) = a + b + c = 1

f (2) = 4 a + 2 b + c = \frac{1}{2}

f (3) = 9 a + 3 b + c = \frac{1}{3}

f (2) - f (1) = 3 a + b = - \frac{1}{2}

3 (f (2) - f (1)) = 9 a + 3 b = - \frac{3}{2}

f (3) - 3 (f (2) - f (1)) = c = \frac{1}{3} - (- \frac{3}{2}) = \frac{11}{6}

f (1) \Rightarrow a + b = 1 - \frac{11}{6} = - \frac{5}{6}

2 a + 2 b = - \frac{5}{3} \dots \dots \dots \dots \dots \dots (i)

f (2) \Rightarrow 4 a + 2 b = \frac{1}{2} - \frac{11}{6} = - \frac{4}{3} \dots \dots (i i)

(i i) - (i i) : 2 a = - \frac{4}{3} - (- \frac{5}{3}) = \frac{1}{3}

a = \frac{1}{6}

f (1) : \frac{1}{6} + b + \frac{11}{6} = 1 \Rightarrow b = 1 - 2 = - 1

f (4) = 16 a + 4 b + c = 16 (\frac{1}{6}) + 4 (- 1) + \frac{11}{6}

= \frac{16 - 24 + 11}{6} = \frac{3}{6} = \frac{1}{2} ✓

Answered by Rasheed.Sindhi last updated on 15/Nov/23

l e t f (x) = {a x}^{2} + b x + c

\begin{array}{ccc} 1) & a & b & c \\ a & a + b \\ a & a + b & f (1) = a + b + c = 1 \end{array}

a = 1 - b - c

\begin{array}{ccc} 2) & 1 - b - c & b & c \\ 2 - 2 b - 2 c & 4 - 2 b - 4 c \\ 1 - b - c & 2 - b - 2 c & f (2) = 4 - 2 b - 3 c = \frac{1}{2} \end{array}

- 4 + 2 b + 3 c = - \frac{1}{2} \Rightarrow b = (\frac{7}{2} - 3 c) / 2 = \frac{7}{4} - \frac{3}{2} c

a = 1 - b - c = 1 - (\frac{7}{4} - \frac{3}{2} c) - c = - \frac{3}{4} + \frac{c}{2}

\begin{array}{ccc} 3) & - \frac{3}{4} + \frac{c}{2} & \frac{7}{4} - \frac{3}{2} c & c \\ - \frac{9}{4} + \frac{3 c}{2} & - \frac{3}{2} \\ - \frac{3}{4} + \frac{c}{2} & - \frac{1}{2} & f (3) = - \frac{3}{2} + c = \frac{1}{3} \end{array}

c = \frac{1}{3} + \frac{3}{2} = \frac{11}{6}

a = - \frac{3}{4} + \frac{c}{2} = - \frac{3}{4} + \frac{1}{2} \cdot \frac{11}{6} = \frac{- 9 + 11}{12} = \frac{2}{12} = \frac{1}{6}

b = \frac{7}{4} - \frac{3}{2} c = \frac{7}{4} - \frac{3}{2} (\frac{11}{6}) = \frac{7 - 11}{4} = - 1

f (x) = \frac{1}{6} x^{2} - x + \frac{11}{6}

f (4) = ?

\begin{array}{ccc} 4) & \frac{1}{6} & - 1 & \frac{11}{6} \\ \frac{2}{3} & - \frac{4}{3} \\ \frac{1}{6} & - \frac{1}{3} & f (4) = \frac{1}{2} \end{array}