f-x-f-x-1-x-1-f-6-solution-

Question Number 194085 by 073 last updated on 27/Jun/23

f (x) - f (x - 1) = x + 1

f^{'} (6) = ?

solution ? ?

Answered by qaz last updated on 27/Jun/23

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

f (x - 1) = a {(x - 1)}^{2} + b (x - 1) + c

f (x) - f (x - 1) = a (2 x - 1) + b = x + 1

\Rightarrow a = \frac{1}{2} b = \frac{3}{2}

\Rightarrow f (x) = \frac{1}{2} x^{2} + \frac{3}{2} x + c \Rightarrow f {(6)}^{'} = \frac{15}{2}

Answered by mr W last updated on 27/Jun/23

a_{k} - a_{k - 1} = k + 1

\underset{k = 1}{\sum^{n}} (a_{k} - a_{k - 1}) = \underset{k = 1}{\sum^{n}} (k + 1)

a_{n} - a_{0} = \frac{n (n + 3)}{2}

a_{n} = \frac{n (n + 3)}{2} + a_{0}

\Rightarrow f (x) = \frac{x (x + 3)}{2} + f (0)

\Rightarrow f^{'} (x) = \frac{2 x + 3}{2}

\Rightarrow f^{'} (6) = \frac{15}{2}

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