Sketch The Domain of Functions
Sketch The Domain of Functions
Sketch The Domain of Functions
√
(a) f (x, y) = x + y + 2
Solution: We need x+y +2 ≥ 0, which is drawn below. The boundary is the line x+y +2 = 0.
-1 -0.5 0 0.5 1
-1
-1.5
-2
-2.5
-3
√ √
(b) f (x, y) = x+ y−3
4.5
3.5
2.5
2
-1 -0.5 0 0.5 1 1.5 2
xy
(c) f (x, y) = x+y−1
0.5
-0.5
-1
-1 -0.5 0 0.5 1
2
(d) f (x, y) = xey +y+1
1.5
0.5
-1
-1.5
-2
Solution: f (3, 2) = 19
ii. f (2, 4, 6)
√
Solution: f (2, 4, 6) = 2 14
Solution: Searching “3D plotter” will yield many options for graphing these.
x3
4. Let f (x) = C (10 − x) for 0 ≤ x ≤ 10 where C is a constant, and f (x) = 0 for all other values of x.
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(a) if f (x) is a probability density function, determine the value of C.
Solution:
∞ 10 10
x3
Z Z Z
1= f (x) dx = (10 − x) dx =⇒ C = x3 (10 − x) dx = 5000
−∞ 0 C 0
(b) For that value of C, find P (1 < X < 4) and P (X > 6) where X is the random variable associated
to f (x).
Solution:
4
x3
Z
4329
P (1 < X < 4) = (10 − x) dx =
1 5000 50000
10
x3
Z
2072
P (X > 6) = (10 − x) dx =
6 5000 3125
It’s not obvious to me how you are expected to solve this equation, but the unique solution in
the inteval [0, 10] is approximately m ≈ 6.86.
5. Given the probability density function f (x) = xe−x for x ≥ 0 and f (x) = 0 for all other values of x.
Find the probability P (X ≥ 1).
Solution: Z ∞
2
P (X ≥ 1) = f (x) dx =
1 e
You can do this integral using integration by parts with u = x and dv = e−x dx.
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