1.5 EXPONENTIAL FUNCTIONS
Lecture 4(25-10-2012)
Exercises (4-14,18-19)
𝒙
4)𝒚 = 𝒆
𝒚=𝒆
−𝒙
𝒚=𝟖
𝒙
𝒚=𝟖
−𝒙
The graph of 𝑒 −𝑥 is the reflection of the graph of
𝑒 𝑥 about the y- axis, and the graph of 8−𝑥 is the
reflection of that 8𝑥 about the y- axis. The graph of
8𝑥 increases more quickly than that for 𝑒 𝑥 for > 0 ,
and approaches 0 faster as 𝑥 → −∞.
Graph For Exercises4
6
𝒚 = 𝟏𝟎𝒙
𝟏
𝒚 = ( )𝒙
𝟑
 y4=8-x
5
 y3=8x
4
3
 y2=e-x
 y1=ex
2
1
0
-2
5)𝒚 = 𝟑𝒙
y
Graph the given functions on a common screen.
How are these graphs related?
-1.5
-1
-0.5
0
0.5
1
1.5
2
x
𝟏
𝒚 = ( )𝒙
𝟏𝟎
Graph For Exercises5
The functions with bases greater than 1(3𝑥 and
10𝑥 ) are increasing, while those with bases less
1
1
3
10
than 1 [( )𝑥 𝑎𝑛𝑑 ( )𝑥 ] are decreasing. The graph
1
of ( )𝑥 is the reflection of 3𝑥 about the y- axis, and
y
6
 y4=(1/10)x
5
 y2=10x
4
 y3=(1/3)x
3
 y1=3x
3
1
the graph of ( )𝑥 is the reflection of that 10𝑥 about
2
10
the y- axis. The graph of10𝑥 increases more quickly
than that for3𝑥 for > 0 , and approaches 0 faster as
𝑥 → −∞.
By:Hessah Alqahtani.
1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
Page 1
2
x
Graph For Exercises6
6)𝒚 = 𝟎. 𝟗𝒙
y
6
 y3=0.3x
5
𝒚 = 𝟎. 𝟔𝒙
 y4=0.1x
4
𝒚 = 𝟎. 𝟑
𝒙
𝒚 = 𝟎. 𝟏
𝒙
3
Each of the graph approaches ∞ 𝑎𝑠 𝑥 → −∞ ,and
each approaches 0 as 𝑥 → ∞. The smaller the
base, the first the function grows as 𝑥 → −∞, and
the faster it approaches 0 as 𝑥 → ∞.
 y2=0.6x
2
1
 y1=0.9x
x
0
-2
-1.5
-1
-0.5
0
0.5
1
7-12 )Make a rough sketch of the graph of the
function. Do not use a calculator. Just use the graph given in Figures 3 and 12 and if
necessary, the transformations of Section 1.3.
7) 𝒚 = 𝟒𝒙 − 𝟑
y
We start with the graph of 𝑦 = 4𝑥 (Figure 3) and then shift 3 units downward. This
shift doesn’t affect the domain, but the range of 𝑦 = 4𝑥 − 3 𝑖𝑠 (−3, ∞). There is a
horizontal asymptote of 𝑦 = −3.
2
2
1
 y=4x
1
y
3
x
0
-1
x
0
 y=(4x )-3
-2
-1
 y=-3
-3
-2
-4
-3
-2
0
2
-5
-2
4
0
2
y=(4x )-3
x
y=4
4
8) 𝒚 = 𝟒𝒙−𝟑
5
4
2
 y=4x
1
3
x
0
2
-1
1
-2
0
-3
-2
0
2
y=4x
By:Hessah Alqahtani.
y
3
y
We start with the graph of of 𝑦 = 4𝑥 (Figure 3) and then shift 3 units to the right. There
is a horizontal asymptote of 𝑦 = 0.
4
-1
-2
 y=4x -3
x
0
2
4
x
y=4 -3
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1.5
2
9) 𝒚 = −𝟐−𝒙
We start with graph of 𝑦 = 2𝑥 (Figure 2), reflect it about the y-axis, and then about the
x-axis(or just rotate 180° to handle both reflections) to obtain the graph of ) 𝑦 = −2−𝑥 .
In each graph, 𝑦 = 0 is horizontal asymptote.
y
Graph For Exercises9
4
4
4
3
3
3
2
2
 y=2-
2
 y=-2-
1
0
1
1
x
 y=2
x
-1
 y=-2-x
-2
0
0
-1
-4
-1
-4
-3
-2
0
2
4
-2
0
2
-4
-4 -2
4
0
2
4
10) 𝒚 = 𝟏 + 𝟐𝒆𝒙
Graph For Exercises10
4
4
3
3
2
2
 y=2ex
1
By:Hessah Alqahtani.
 y=1+2.ex
1
0
-1
-4
y
We start with the graph of 𝑦 = 𝑒 𝑥 (Figure 13), vertically stretch by a factor of 2, and
shift 1 unit upward. There is a horizontal asymptote of 𝑦 = 1.
y=1
0
-2
0
2
4
-1
-4
x
-2
0
2
4
Page 3
𝟏
11) 𝒚 = 𝟏 − 𝒆−𝒙
𝟐
We start with the graph of 𝑦 = 𝑒 𝑥 (Figure 13), and reflect about the y-axis to get the
graph of = 𝑒 −𝑥 . Then we compress the graph vertically by a factor of 2 to obtain the
1
1
graph of 𝑦 = 𝑒 −𝑥 and then reflect about the x-axis to get the graph of 𝑦 = − 𝑒 −𝑥 .
2
2
1
Finally, we shift the graph upward one unit to get the graph of 𝑦 = 1 − 𝑒
−𝑥
2
Graph For Exercises11
4
4
4
4
3
3
3
2
2
2
2
2
 y=1/2*e-x 0
0
1
x
 y=e 1
-
 y=e x 1
0
0
-1
-4 -2 0
2
y=1
y
y
4
-1
4 -4 -2 0 2
0
4
-1
-4 -2 0 2 4
x
-2
-4
-4
.
 y=(-1/2)e(-x)
-2
0
12) 𝒚 = 𝟐(𝟏 − 𝒆𝒙 )
2
-4
-4
4
-2
0
x
2
4
Graph For Exercises12
4
4
We start with the graph of 𝑦 = 𝑒 𝑥 (Figure 13), and
reflect about the x-axis to get the graph of = 𝑒 −𝑥 .
Then shift the graph upward one unit to get the
graph of 𝑦 = (1 − 𝑒 𝑥 ). Finally, we stretch the
graph vertically by a factor of 2 to obtain the
graph of 𝑦 = 2(1 − 𝑒 𝑥 ).
 y=1+(-1/2)e(-x)
-2
3
2
 y=ex
2
0
1
 y=-ex
-2
0
-1
-4
-2
0
2
-4
-4
4
4
-2
0
2
x
 y=2(1-ex )
y
 y=1-e
By:Hessah Alqahtani.
4
4
2
0
0
-2
-2
-4
-4
2
-2
0
x
2
4
-4
-4
-2
0
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2
4
13) Starting with the graph of 𝒚 = 𝒆𝒙 , writ the equation of the graph that results
from
a)shifting 2 units downward
To find the equation of the graph that results from shifting the graph of 𝑦 = 𝑒 𝑥 2 units
downward, we subtract 2 from the original function to get 𝑦 = 𝑒 𝑥 − 2.
b) shifting 2 units to the right
To find the equation of the graph that results from shifting the graph of 𝑦 = 𝑒 𝑥 2units
to the right, we replace x with x- 2 in the original function to get 𝑦 = 𝑒 (𝑥−2) .
c) reflecting about the x-axis
To find the equation of the graph that results from reflecting the graph of 𝑦 = 𝑒 𝑥 about
the x-axis, we multiply the original function by -1 to get 𝑦 = −𝑒 𝑥
d) reflecting about the y-axis
To find the equation of the graph that results from reflecting the graph of 𝑦 = 𝑒 𝑥 about
the y-axis, we replace x with-x in the original function to get 𝑦 = 𝑒 −𝑥
e) reflecting about the x-axis and then about the y-axis
To find the equation of the graph that results from reflecting the graph of 𝑦 = 𝑒 𝑥 about
the x-axis and then about the y-axis, we first multiply the original function by -1 to get
𝑦 = −𝑒 𝑥 , and then replace x with-x in this equation to get 𝑦 = −𝑒 −𝑥 .
By:Hessah Alqahtani.
Page 5
14) Starting with the graph of 𝒚 = 𝒆𝒙 , find the equation of the graph that results
from
a) reflecting about the line y=4
This reflection consists of the first reflecting the graph about the x-axis(giving the graph
with equation 𝑦 = −𝑒 𝑥 ) and then shifting this graph 2.4=8 units upward. So the
equation is 𝑦 = −𝑒 𝑥 + 8.
b) reflecting about the line x=2
This reflection consists of the first reflecting the graph about the y-axis(giving the graph
with equation 𝑦 = 𝑒 −𝑥 ) and then shifting this graph 2.2=4 units to the right. So the
equation is 𝑦 = 𝑒 −(𝑥−4) .
18) Find the exponential function 𝒇(𝒙) = 𝑪𝒂𝒙 whose graph is given.
2
Given the y-intercept (0,2)we have 𝑦 = 𝐶𝑎 𝑥 = 2𝑎 𝑥 . Using the point (2, ) gives us
9
2
9
1
1
9
3
= 2𝑎2 → = 𝑎2 → 𝑎 =
[since 𝑎 > 0]. The function is
1 𝑥
𝑓 (𝑥) = 2 ( ) 𝑜𝑟 𝑓(𝑥) = 2(3)−𝑥 .
3
19) If 𝒇(𝒙) = 𝟓𝒙 , show that
If 𝑓 (𝑥) = 5𝑥 , then
By:Hessah Alqahtani.
𝒇(𝒙+𝒉)−𝒇(𝒙)
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ
𝒉
=
= 𝟓𝒙 (
5𝑥+ℎ −5𝑥
ℎ
=
𝟓𝒉 −𝟏
𝒉
)
5𝑥 5ℎ −5𝑥
ℎ
=
5𝑥 (5ℎ −1)
ℎ
= 5𝑥 (
5ℎ −1
ℎ
)
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