30 Higher Engineering Mathematics
x −3.0 −2.5 −2.0 −1.5 −1.0 −0.5 0
e x
0.05 0.08 0.14 0.22 0.37 0.61 1.00
e −x 20.09 12.18 7.39 4.48 2.72 1.65 1.00
x
0.5
1.0
1.5
2.0
2.5
3.0
e x
1.65 2.72 4.48 7.39 12.18 20.09
e −x 0.61 0.37 0.22 0.14
0.08
0.05
Figure 4.1 shows graphs of y = e x and y = e −x
y
20
16
y 5 e x
y 5 e 2x
12
8
4
21
0
1
2
3
x
22
23
Figure 4.1
Problem 6. Plot a graph of y = 2 e 0.3x over a
range of x =−2 to x = 3. Hence determine the value
of y when x = 2.2 and the value of x when y = 1.6.
A table of values is drawn up as shown below.
x
−3 −2 −1
0
1
2
3
0.3x −0.9 −0.6 −0.3 0
0.3 0.6 0.9
e 0.3x 0.407 0.549 0.741 1.000 1.350 1.822 2.460
2 e 0.3x 0.81 1.10 1.48 2.00 2.70 3.64 4.92
A graph of y = 2 e 0.3x is shown plotted in Fig. 4.2.
From the graph, when x = 2.2, y = 3.87 and when
y = 1.6, x =−0.74.
y
5
y5 2e 0.3x
4
3
1.6
3.87
1
20.74
2.2
21
0
1
2
3
x
22
23
2
Figure 4.2
Problem 7. Plot a graph of y =
1
3 e −2x over the
range x =−1.5 to x = 1.5. Determine from the
graph the value of y when x =−1.2 and the value
of x when y = 1.4.
A table of values is drawn up as shown below.
x
−1.5 −1.0 −0.5 0 0.5 1.0 1.5
−2x
3
2
1
0 −1 −2 −3
e −2x 20.086 7.389 2.718 1.00 0.368 0.135 0.050
1
3
e −2x 6.70 2.46 0.91 0.33 0.12 0.05 0.02
A graph of
1
3 e −2x is shown in Fig. 4.3.
7
6
5
4
3
3.67
1.4
2
1
0.5
20.5
20.72
21.0
21.2
21.5
1.0
1.5
y
x
1
3
e
22x
y 5
Figure 4.3
From the graph, when x =−1.2, y = 3.67 and when
y = 1.4, x =−0.72.
Problem 8. The decay of voltage, v volts, across
a capacitor at time t seconds is given by
v = 250 e
−t
3 . Draw a graph showing the natural