S o l u t i o n : E x a m p l e 1 ) L e t f ( x ) = x 2 + s i n ( x ) − 1 T o a p p l y t h e I n t e r m e d i a t e V a l u e T h e o r e m t o t h e e q u a t i o n f ( x ) = 0 o n t h e i n t e r v a l [ 0 , π ] , w e e v a l u a t e f ( 0 ) a n d f ( π ) f ( 0 ) = 0 2 + s i n ( 0 ) − 1 = − 1 < 0 a n d f ( π ) = π 2 + s i n ( π ) − 1 = π 2 − 1 > 0 T h e n b y I n t e r m e d i a t e V a l u e T h e o r e m , t h e r e i s a s o l u t i o n o f f ( x ) = 0 i n t h e i n t e r v a l [ 0 , π ] . T h u s w e c o n c l u d e t h a t t h e r e i s a r o o t o f t h e f u n c t i o n f ( x ) = x 2 + s i n ( x ) − 1 i n t h e i n t e r v a l [ 0 , π ] E x a m p l e 2 ) : L e t f ( x ) = 1 x 1 − x 2 d e f i n e d o n − 1 < x < 1 I t s v a l u e s a r e n e g a t i v e f o r n e g a t i v e x , p o s i t i v e x , b u t n e v e r z e r o . E x a m p l e 3 ) : L e t f ( x ) = 1 i f − 1 ≤ x ≤ 0 a n d f ( x ) = − 1 i f 0 < x ≤ 1 H e r e f ( x ) h a s o p p o s i t e s i g n s a t t h e e n d p o i n t s w h i l e i t i s n e v e r z e r o i n t e i n t e r v a l . Solution:
\\ Example~1)
\\ Let~ f(x)=x^2+sin(x)-1
\\To~apply~the ~Intermediate~Value~Theorem~to~the~equation~f(x)=0~on~the~interval~[0,\pi], we ~evaluate~f(0) ~and~ f(\pi)
\\f(0)=0^2 +sin(0)-1=-1<0~and~f(\pi)=\pi^2+sin(\pi)-1=\pi^2-1>0
\\Then~ by ~ Intermediate~Value~Theorem,there ~is ~ a~ solution~of~f(x)=0
\\in~the ~interval~[0,\pi].
\\Thus~ we ~conclude ~that~there~is~a ~root~of~the~function~f(x)=x^2+sin(x)-1~in~the~interval~[0,\pi]
\\
\\Example ~2):
\\Let~f(x)=\frac{1}{x\sqrt{1-x^2}} defined on -1<x<1
\\Its ~values ~are~negative~for~negative~x,positive~x,but~never ~zero.
\\Example~3) :
\\Let~f(x)=1~if~-1 \le x \le 0~and f(x)=-1~if ~0<x \le1
\\Here~f(x)~has ~opposite~signs~at~the~end~points~while~it~is~never~zero~in~te~interval. S o l u t i o n : E x am pl e 1 ) L e t f ( x ) = x 2 + s in ( x ) − 1 T o a ppl y t h e I n t er m e d ia t e Va l u e T h eore m t o t h e e q u a t i o n f ( x ) = 0 o n t h e in t er v a l [ 0 , π ] , w e e v a l u a t e f ( 0 ) an d f ( π ) f ( 0 ) = 0 2 + s in ( 0 ) − 1 = − 1 < 0 an d f ( π ) = π 2 + s in ( π ) − 1 = π 2 − 1 > 0 T h e n b y I n t er m e d ia t e Va l u e T h eore m , t h ere i s a so l u t i o n o f f ( x ) = 0 in t h e in t er v a l [ 0 , π ] . T h u s w e co n c l u d e t ha t t h ere i s a roo t o f t h e f u n c t i o n f ( x ) = x 2 + s in ( x ) − 1 in t h e in t er v a l [ 0 , π ] E x am pl e 2 ) : L e t f ( x ) = x 1 − x 2 1 d e f in e d o n − 1 < x < 1 I t s v a l u es a re n e g a t i v e f or n e g a t i v e x , p os i t i v e x , b u t n e v er zero . E x am pl e 3 ) : L e t f ( x ) = 1 i f − 1 ≤ x ≤ 0 an df ( x ) = − 1 i f 0 < x ≤ 1 Here f ( x ) ha s o pp os i t e s i g n s a t t h e e n d p o in t s w hi l e i t i s n e v er zero in t e in t er v a l .
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