Here -2 lies in , 1 lies in , 2 lies in , 13 lies in .
Now,
i.e.,
And,
i.e.,
And,
i.e.,
And,
i.e.,
Thereore, for x = -2 and x = 2, we have f(x) = 4.
consider continuous joint density function f(x,y)= (x+y)/7; 1<x<2, 1<y<3 Marginal density for Y? Select one: (2+3x)/14 (3+2y)/7 (2+3y)/14 (3+2y)/14 consider continuous joint density function f(x,y)= (x+y)/7 ; 1<x<2, 1<y<3 P(0<x<3, 0<y<4)=? Select one: 0.5 1 0.15 0.25
1. Given the piece-wise function, 3x if x < 0 f(x)=x+1 if 0 < x 52 :- 2)2 if x>2 Evaluate f (__); f(0); f (); f(5)
solve for c such that f(x,y) is a valid density function. Seiten f(x, y) = 1<x<y <3 otherwise 0,
(1 point) Find the maximum and minimum values of the function f(x, y) = 3x² – 18xy + 3y2 + 6 on the disk x2 + y2 < 16. Maximum = Minimum =
Suppose that the functionſ is defined, for all real numbers, as follows. 3x+1 fx < -2 x-3 if x 2-2 Graph the functionſ. Then determine whether or not the function is continuous. Is the function continuous? 10 X o Yes NO O X ? 2 8 10 -10 Continue
6. If 3x – 3 5 f(x) < x2 – 3x + 6 for x 2 0, find lim f(x). X-3
Given the following piecewise function, evaluate f(-5). I < -4 f(x) = . 1-42 -3x (x² – 2 -4 < x < 0 0<x
* For the function f(x) = (3x + 5 if x 20 6 if x < 0. find f(-10)
Find all values x = a where the function is discontinuous. 3x - 5 if x < 0 f(x) = x2 + 5x -5 if x 20 O A. a = 0 OB. Nowhere O c. a = 5 OD. a = -5
Let g be the function defined by 1 if x < 6 g(x) = 2 - 6 if x 26. Find g(-6), g(0), g(6), and g(12). 9(-6) g(0) = 9(6) 9(12) =