Problem :
What is
x
2 + 3x - 5
?
x
2 + 3x - 5 = 5
(direct substitution).
Problem :
What is
?
= 0
. By factoring
x
out of the numerator and
denominator,
f
is defined at
x = 0
, and direct substitution works.
Problem :
What is
4
?
4 = 4
. This is a constant function.
Problem :
What is
?
=
= 
.
Problem :
Let
f (x) = x
for
x < 1
,
f (x) = 1
for
x≥1
. What is
f (x)
?
f (x) = 1
.
Problem :
Let
f (x) = x
2
for
x < 0
,
f (x) = - 1
for
x≥ 0
. What are
f (x)
,
f (x)
, and
f (x)
?
f (x) = 0
.
f (x) = - 1
.
f (x)
doesn't exist, because
f (x)≠
f (x)
.
