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, ∞ )