The function is defined over these ranges of x:
(-∞, -1) and [-1, 7), that is for x < 7.
Note that
The function is undefined at x = --1 over the left portion of the straight line, but the function s defined at x = -1 for the right portion f the straight line.
The function is undefined at x = 7 and for larger values of x.
Answer: x < 7
Answer:
The Martian satellite Phobos travels in an approximately circular orbit of radius 9.4x10^6m with a period of 7 h39 min. Calculate the mass of Mars from this information.
Step-by-step explanation:
Here we have
f
(
x
)
=
2
x
2
(
x
2
−
9
)
, which can be factorized as
f
(
x
)
=
2
x
2
(
x
+
3
)
(
x
−
3
)
As there is no common factor between numerator and denominator, there s no hole.
Further vertical asymptotes are
x
=
−
3
and
x
=
3
and as
f
(
x
)
=
2
x
2
(
x
2
−
9
)
=
2
1
−
9
x
2
, as
x
→
∞
,
f
(
x
)
→
2
, hence horizontal asymptote is
y
=
2
.
Observe that
f
(
−
x
)
=
f
(
x
)
and hence graph is symmetric w.r.t.
y
-axis. Further as
x
=
0
,
f
(
x
)
=
0
. Using calculas we can find that at
(
0
,
0
)
there is a local maxima as
d
y
d
x
=
−
36
x
(
x
2
−
9
)
2
and at
x
=
0
it is
0
. Further while for
x
<
−
3
and
x
>
3
, function is positive, for
−
3
<
x
<
3
function is negative.
Now take a few values of
x
say
{
−
10
,
−
7
,
−
4
,
−
2
,
−
1
,
1
,
2
,
4
,
7
,
10
}
and corresponding values of
f
(
x
)
are
{
2
18
91
,
2
9
20
,
4
4
7
,
−
1
3
5
,
−
1
4
,
−
1
4
,
−
1
3
5
,
4
4
7
,
2
9
20
,
2
18
91
}
0.6 (y + 3) = 4.8
y+3 =4.8/0.6
y + 3 = 8
y = 8 - 3 = 5
y = 5
Answer:
a) P ( 3 ≤X≤ 5 ) = 0.02619
b) E(X) = 1
Step-by-step explanation:
Given:
- The CDF of a random variable X = { 0 , 1 , 2 , 3 , .... } is given as:
Find:
a.Calculate the probability that 3 ≤X≤ 5
b) Find the expected value of X, E(X), using the fact that. (Hint: You will have to evaluate an infinite sum, but that will be easy to do if you notice that
Solution:
- The CDF gives the probability of (X < x) for any value of x. So to compute the P ( 3 ≤X≤ 5 ) we will set the limits.
- The Expected Value can be determined by sum to infinity of CDF:
E(X) = Σ ( 1 - F(X) )
E(X) = Limit n->∞ [1 - 1 / ( n + 2 ) ]
E(X) = 1
