Solución de la práctica dirigida 6

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C L E G I O
IJ K C ] Y E C rJ
M A T L:M A T IC A
I'
H C T IC A D IR IG M N 6
s o b c z o hu a R o D E s e C U N D A m A "
F IR M A D E L P A D R E O A P O D E R A D O
S in h h r o s n i a p u n te s
N O T A D c b c r tis e s c r ib ir la s r e s p u e s ta s c o n la p ic e r o
P R O Y B C T O W 1 D a d o s lo s c o n ju n to s
B { 2 ; 0 ; 5 }
ic u M d e la s s ig u ie n te s fu n c io n e s n o E s u n a Fu n c ió n d e B e n A ?
A f 2 Y U U 4 V
W 2 2 3 H 7 0 f
e J u c t o j
P R O Y E C T O N 2
in d ic a r e l v a lo r n u m éric o d e a b
S i e l c o n ju n to : {( 5 ; a + n (Z b 7 ) ( 9 ).
( 5 ; " } e s u n a filn c ió n
c t b ・ q l
P R O l T O N
P R O Y E C ? O N 4 S i fi(x ) 4 x I; g (x ) 2 x + 13 h a lla r f(8 ( 4
p R o ￥E c T o N °
s D a d o : f(x ) 3x 1 C a lc u la r : f(2 ) fil2 )
f n ). 5 f w - 1
S + 1
C t H = l0 b 7 r
￡
)
C l r 9 b = l

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IO}+ !W )
P R O Y E C T O W 8 S i s e tie n e n la s Ih n c io n e s : f(x ) 8×
2
5 ; g (x ) ×
3
3 C a lc u la r : fi(3)+ g j 2 )
p n o E m o w 9 D a d a s a s hmc ion es f W 2 3 Y 5 30 g 3 2 1 5 2 3 11
h a n a r
m 3 k z
H s r H d 2
" -
+-
Yf (s )+ ffa
)
D e Ia p r e g u n ta 10 a la 13 J g r a fic a r c a d a u n a d e la s fu n c io n e s lin e a le s h a lla r e l d o m in io y
r a n g o
P R O ￥ E C T O W 1 0 y 2 x + 4 X y r 2 x + tt
D o m ・ R
R a n g - IL

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