Sample Graph Sketching Problem - ms.mcmaster.camcleac3/math1ZA/Notes/ExpGraphing.pdf · Sample...
Transcript of Sample Graph Sketching Problem - ms.mcmaster.camcleac3/math1ZA/Notes/ExpGraphing.pdf · Sample...
Sample Graph Sketching Problem1Let'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 arecontinuous 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
xxx
x
x x x
==
≠ ( ) ( )
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 exponentialx
x
y e
=
= =
:
:
y - intercept
x - intercept
1 1
1 1
0
0
s are > 0.
Notice: lim " " 1 HA: 1 as
lim " " 1 HA: 1 as
So we have t
x
x
x
x
e e e y x
e e e y x
∞
− ∞
→ ∞
→ − ∞
= = = ⇒ = → ∞
= = = ⇒ = → − ∞
Step #4 : Asymptotes:Horizontal Asymptotes ("HA")
he same 1 horizontal asymptote in both directions.
These are only going to occur at possible discontinuities, so the only place they exist on our g
y
could
=
:Vertical Asymptotes ("VA")
1 11 10 0
0 0
raph is at 0.
lim " " " " , but lim " " " " 0
So we do have a vertical asymptote, but as we approach from the right!
x x
x x
x
e e e e e e
ONLY
+ −
+ −
+ ∞ − ∞
→ →
=
= = = ∞ = = =
11 1 1
2 2
2
1 1( ) '( )
Notice, that exponentials are all positve, as is , so '( ) < 0 ( )
xx x x
where it's defined...
d ef x e f x e edx x x
x f x
= ⇒ = × = × = −
− x
Step #5 : Examine '( )f x
This means our function is decreasing everywhere, and there are no critical points.
But let's grind this through formally for the practice.
:Critical PointsCa
11
2
0
'( ) 0 0
But again, real exponentials are always positve, so this never happens.
xx
ef x ex
=
= − = ⇒ =
: '( )
: '( ) = DNE,
se#1 f x
Case#2 f x (but
1
2
)
'( ) 0, but that's not in the domain.
So we have no critical points (c.p.'s) whatsoever. But we do have a gap at zero, so
xef x DNE xx
= − = ⇒ =
( )f x exits.
when we look for the intervals of increasing and decreasing, we'll still keep 0 in mind.
By inspection, as discussed above, or by plug
x =
:Intervals of Increasing and Decreasingging in a sample -value from each
interval, we can find the sign on each interval:
( ,0) (0, ) '( ) " " " " ( )
x
f xf x decreasing d
− ∞ ∞− −
Normally we could use the information from the chart above to do a first derivative test. But no c.p.'s means
ecreasing
no local maximum or minim
:Local maximums and mimimums
either. um
( )
11
11 1 1
1
22
2 22
2 42
'( )
2''( )
xx
xx x
ef x x ex d e
e x × x + 2xe xedx xf xxx
e
−= − = −
× x + ⇒ = =
=
Step #6 : Examine ''( )f x
( )
( )
1 1
1
4 4
4
4
2 1 2
Again, exponentials are all positve, as is .
0 ( )
and 1 2 , which changes sign.
x x x
x
again, where it's defined...
xe e xx x
x
ex
x
− = +
We, then, have two factors: >
+
As before, we look for these as we did the critical points, but using the 2nd derivative. 0=
:
: ''( )
Inflection Points (i.p.'s)
Case#1 f x
( ) ( )
( )
1
1
4
4
1 f "(x) 1 2 0 1 2 0 2
)
) 1 2 0, but, as before, that's not in
x
x
e x x xx
e f "(x x DNE xx
−= + = ⇒ + = ⇒ =
= + = ⇒ =
: ''( ) = DNE, ( )Case#2 f x (but f x exits.
the domain.1 This means we have one inflection point, the one at .2
We'll do the concavity chart, and a
possible x −=
t the same time we can verify if this is, in fact an i.p.
1 We use both our potential i.p. at and our discontinuity to set our boundries.2x −=
:Intervals of Concavity
1
1 12 2
4
This gives us the intervals: ( , ), ( ,0), (0, )
Let's examine f "(x) on each interval: ( , 0 )xeand remember
x
− ∞ − − ∞
>
-
( ) ( )
( ) ( )
( ) ( )
1
1
1
12 4
12 4
4
( , ) 1 2 0 1 2 0
( ,0) 1 2 0 1 2 0
( 0, ) 1 2 0 1 2 0
And we get a
x
x
x
ex x xxex x xxex x xx
∈ − ∞ − ⇒ + < ⇒ + <
∈ − ⇒ + > ⇒ + >
∈ ∞ ⇒ + > ⇒ + >
chart:
) " − "
( ) concave down
" + "
concave up
(C.D.) (C.U.)
" + "
concave up
(C.U.)
f "(x
f x
1 Notice, that at , we get a . So yes, indeed it is an i.p.2
x change in concavity−=
Step #7 : Graph It
1 12 2 (− ∞ ,− ) (− ,0) (0, ∞ )