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Question Number 73567 by Rio Michael last updated on 13/Nov/19

$D e t e r m i n e t h e v a l u e o f a a n d b f i r w h i c h t h e f u n c t i o n f, d e f i n e d b y$ $f (x) = {\begin{cases} - 2 s i n x, x < - \frac{π}{2} \\ a s i n x + b, - \frac{π}{2} ⩽ x < \frac{π}{2} \\ c o s x, x > \frac{π}{2} \end{cases}$ $i s c o n t i n o u o s$
Commented by Rio Michael last updated on 13/Nov/19

$t h a n k y o u$
Commented by kaivan.ahmadi last updated on 13/Nov/19
$lim_(x→−(π^− /2)) f(x)=−2sin(−(π/2))=2 lim_(x→−(π^+ /2) ) f(x)=lim_(x→−(π^+ /2)) (asinx+b)=−a+b ⇒−a+b=2 and similarly for x=(π/2) we have a+b=0 ⇒ { ((−a+b=2)),((a+b=0)) :}⇒a=−1,b=1.$
${l i m}_{x \to - \frac{π^{-}}{2}} f (x) = - 2 s i n (- \frac{π}{2}) = 2$ ${l i m}_{x \to - \frac{π^{+}}{2}} f (x) = {l i m}_{x \to - \frac{π^{+}}{2}} (a s i n x + b) = - a + b$ $\Rightarrow - a + b = 2$ $a n d s i m i l a r l y f o r x = \frac{π}{2} w e h a v e$ $a + b = 0$ $\Rightarrow {\begin{cases} - a + b = 2 \\ a + b = 0 \end{cases} \Rightarrow a = - 1, b = 1 .$
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