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Question Number 67684 by Rio Michael last updated on 30/Aug/19

g i v e n t h e f u n c t i o n

f (x) = {\begin{cases} x^{2}, f o r 0 ⩽ x < 2 \\ a x + 3, f o r 2 ⩽ x < 4 \end{cases}

i s p e r i o d i c o f p e r i o d 4, a n d i s c o n t i n u o u s .

a) F i n d t h e v a l u e o f a .

b) F i n d t h e v a l u o f f (6)

c) s k e t c h t h e g r a p h f o r y = f (x) .

h e l p m e p l e a s e, f o r t h e g r a p h i {d o n}^{'} t k n o w w b e r e t o p u t y = x^{2} a n d y = a x + 3 a n d

w h e r e d o i p u t a c l o s e d d o t a n d a n o p e n d o t .

Commented by MJS last updated on 30/Aug/19

if f (x) is continuous within [0; 4] \Rightarrow

\Rightarrow x^{2} = a x + 3 for x = 2 \Rightarrow a = \frac{1}{2}

but then {it}^{'} s not continuous at 4 ({it}^{'} s periodic \Rightarrow

\Rightarrow it “ starts {again}^{″} at x = 4 with the same

value as at x = 0

f (0) = 0

\lim_{x \to 4^{-}} f (x) = \frac{1}{2} \times 4 + 3 = 5

\lim_{x \to 4^{+}} f (x) = 0^{2} = 0

so {it}^{'} s not continuous \dots

anyway the value of f (6) = f (6 - 4) = f (2) = 4

closed dots at ⩽ and ⩾

open dots at < and >

Commented by Rio Michael last updated on 14/Sep/19

t h a n k y o u s i r, s o m u c h