Sample Graph Sketching Problem
1
Let's do a graph sketching problem.
We'll look at the graph of the function:
( )
It's probably easier to find what isn't in the domain. Functions like
are
continuous on all
x
x
y
e
f x
e
=
=
Step #1: Domain
{
}
1
. The only possible problem here is with
.
This function is undefined at
0, so our composite function will also be
undefined at
0.
Our domain then is:
0 , or in other words, all
such t
x
x
x
x
x x
x
=
=
≠
(
29
(
29
1
1
(
)
hat
,0
0,
Let's check the usual suspects.
1)
: Does
( )
(
) for some real
value?
2)
/
: We look at
(
)
,
x
x
x
Periodic
f x
f x
a
a
NO
Even Odd
f
x
e
e


∈
 ∞
∪
∞
=
+

=
=
Step #2: Symmetry
1
1
(
)
(
), so not odd
( ), so not even
In all, no usable symmetries.
Only bother with these i
x
x
f
x
e
f
x
e
f x



=
≠

≠
Step #3: Intercepts
1
f they are reasonably easy to calculate.
Here, they are especially easy:
0 isn't in the domain, so no intercept here.
0
, that never happens. Real exponential
x
x
y
e
=
=
=
:
:
y  intercept
x  intercept
1
1
1
1
0
0
s are > 0.
Notice: lim
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 Spring '11
 CHILDS
 Calculus, Derivative, Mathematical analysis, Limit of a function, ex

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