f-x-y-f-x-1-y-x-1-f-x-y-ye-x-x-0-f-5-6-

Question Number 2604 by 123456 last updated on 23/Nov/15

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

f (x, y) = {y e}^{x}, x ⩽ 0

f (5, 6) = ?

Answered by Yozzis last updated on 23/Nov/15

f (5, 6) = f (4, 1) + 1

f (4, 1) = f (3, - 3) + 1 \Rightarrow f (5, 6) = f (3, - 3) + 2

f (3, - 3) = f (2, - 6) + 1 \Rightarrow f (5, 6) = f (2, - 6) + 3

f (2, - 6) = f (1, - 8) + 1 \Rightarrow f (5, 6) = f (1, - 8) + 4

f (1, - 8) = f (0, - 9) + 1 \Rightarrow f (5, 6) = f (0, - 9) + 5

I n f (0, - 9), x = 0 a n d y = - 9

∴ f (0, - 9) = - 9 \times e^{0} = - 9 (p o s s i b l y)

f o r x - 1 ⩽ 0, f (x, y) = f (x - 1, y - x) + 1 = (y - x) e^{x - 1} + 1

∴ f (0, - 9) = (- 9 - 0) e^{- 1} + 1 = 1 - 9 e^{- 1}

∴ f (5, 6) = 5 - 9 = - 4 (p o s s i b l y) o r f (5, 6) = 6 - 9 e^{- 1}

f (5, 6) t a k e s i n f i n i t e l y m a n y v a l u e s b y t h e d e f i n i t o n

o f f (x, y) = f (x - 1, y - x) + 1 .

I f f (x, y) = f (x - 1, y - x) + 1 f o r x ⩾ 1 o n l y t h e n f (5, 6) = - 4 .

Commented by prakash jain last updated on 23/Nov/15

The font looks really small .

Commented by Yozzis last updated on 23/Nov/15

I {d o n}^{'} t k n o w w h y t h i s h a s h a p p e n e d \dots

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