let f(x)=│x-3│ which of the following conditional statement hold true
A.f(x)=-(x-3)
B.f(x)0 if x=0
C.f(x)=x-3 if x>0
D.f(x)=x-3 if x>3
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OpenStudy (anonymous):
@satellite73 can u help
OpenStudy (sidsiddhartha):
if f(x)=|x|
then
f(x)=x ,when x>0
f(x)=-x,when x<0
use this property
OpenStudy (anonymous):
can u help me solve it
OpenStudy (unklerhaukus):
consider x is 5
is A correct?
OpenStudy (anonymous):
yea
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OpenStudy (unklerhaukus):
check that
f(x)=│x-3│
f(5)=│5-3│
= ?
A. f(x)=-(x-3)
f(5)=-(5-3)
= ??
are ? and ?? equal
OpenStudy (anonymous):
ohh
OpenStudy (unklerhaukus):
what do you get for ? and ??
OpenStudy (anonymous):
im trying to figure it out
OpenStudy (anonymous):
can u help me
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OpenStudy (unklerhaukus):
f(5)=│5-3│
what is 5-3
OpenStudy (anonymous):
2
OpenStudy (unklerhaukus):
f(5)=│5-3│
= | 2|
what is absolute value of 2
OpenStudy (anonymous):
2
OpenStudy (unklerhaukus):
good so
f(x)=│x-3│
f(5)=│5-3│
= | 2 |
= 2
A. f(x) = -(x-3)
f(5) = -(5-3)
= -(2)
are 2 and -(2) equal? could A. be correct?
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OpenStudy (anonymous):
no
OpenStudy (unklerhaukus):
good, so we can rule out option A.
OpenStudy (anonymous):
ok
OpenStudy (unklerhaukus):
lets look at option B. now
f(x)=│x-3│
B. f(x) = 0 if x=0
lets compute f(0)
f(0) =│0-3│
=
what does this simplify to become?
OpenStudy (anonymous):
3
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OpenStudy (unklerhaukus):
thats right f(0) = 3 is this that same as B. f(x) = 0 if x=0 ?
OpenStudy (anonymous):
yea
OpenStudy (unklerhaukus):
3 is the same as 0?
OpenStudy (anonymous):
oh noo
OpenStudy (unklerhaukus):
3 and 0 are not the same are they, so can option B. be the correct one?
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OpenStudy (anonymous):
no
OpenStudy (unklerhaukus):
f(x)=│x-3│
How about option C.
C. f(x) = x-3 if x>0
lets try x = 7 because 7>0
what is f(7) =
OpenStudy (anonymous):
7
OpenStudy (unklerhaukus):
f(7) =│7-3│
= . . .
OpenStudy (anonymous):
4
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OpenStudy (unklerhaukus):
ok and
C. f(7) = 7-3
= . . .
OpenStudy (anonymous):
4
OpenStudy (unklerhaukus):
that's right so we tested x=7 and found no problem with C.
but what if we tested x = 2 because 2 is still greater than zero
f(x) =│x-3│
f(2) = | 2-3 |
=
C. f(x) = x - 3
f(2) = 2 - 3
=
are these the same?
OpenStudy (anonymous):
yes
OpenStudy (unklerhaukus):
what did you get for
f(2) = | 2-3 |
=
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OpenStudy (anonymous):
1
OpenStudy (unklerhaukus):
yeah,
and what about
C. f(x) = x - 3
f(2) = 2 - 3
=
OpenStudy (anonymous):
so c is correct
OpenStudy (unklerhaukus):
what did you get for 2 - 3
OpenStudy (anonymous):
-1
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OpenStudy (anonymous):
so it has to Be D
OpenStudy (unklerhaukus):
is 1 the same as -1 ?
OpenStudy (anonymous):
no
OpenStudy (unklerhaukus):
so now lets check D.
f(x)=│x-3│
D. f(x)=x-3 if x>3
what is f(4) ?
OpenStudy (anonymous):
4
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OpenStudy (anonymous):
4-3=1
OpenStudy (anonymous):
│4-3│=1
OpenStudy (unklerhaukus):
that fits with
f(x)=│x-3│
and with
D. f(x)=x-3 if x>3
what about f(23)?
OpenStudy (anonymous):
│23-3│=20
OpenStudy (unklerhaukus):
and
D. f(x) = 23-3 (23>3)
= 20 also
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OpenStudy (anonymous):
so its D
OpenStudy (unklerhaukus):
yep
\[\checkmark\]
OpenStudy (anonymous):
can u help me with1 more
OpenStudy (unklerhaukus):
i think i have to go to bed, soory
OpenStudy (anonymous):
it wont take long plx i just beed anwser
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OpenStudy (unklerhaukus):
go on then
OpenStudy (anonymous):
let f(x)= -2x x<0
0 x>0
which is equivalent
A.f(x)=x+│x│
B.f(x)=│x│-x
C.f(x)=2x
D.f(x)=2│x│
OpenStudy (unklerhaukus):
test some numbers for x
OpenStudy (anonymous):
what do u think the anwser is
OpenStudy (unklerhaukus):
what if x=1 ? (1>0)
is A. correct?
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OpenStudy (anonymous):
yea
OpenStudy (anonymous):
wait no
OpenStudy (unklerhaukus):
can you show some working
OpenStudy (anonymous):
i only have 5 min to finish and this is the last question
OpenStudy (unklerhaukus):
im not going to tell you the answer, but i can check your working
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