12. Suppose f(x) is defined as shown below +3 if s 2 f(x) = 3x if...
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
Suppose that the piecewise function J is defined by f(2)= {**** -1<<3 - 3x2 + 2x + 23, 2> 3 Determine which of the following statements are true. Select the correct answer below: O f() is not continuous at I = 3 because it is not defined at I = 3. Of() is not continuous at 2 = 3 because lim f(x) does not exist. f() is not continuous at I = 3 because lim f() f(3). ->3 f(x) is...
A function is defined as follows: y = X + 6 x² 3x + 1 X<-2 -2<x<3 x > 3 For which x-values is f(x) = 4? Select all that apply 0-2 1 2. 13 e here to search
find the limit f(x) = 2x2 -3, if x-2 13-x, ifx < -2 then, determine if the function is continuous at x=-2
Suppose f'(x) = -1/3(x+3). On what open interval(s) is f(x) decreasing 0-3 < < 0 0-3 <x<0 0 - < I< -3 0 - < ?<-3 and 0 < x < 0
cewise Functions e function, evaluate lim f(x). 2 1-2x²+x+3 f(x) = { 2x2 – 3x + 3 (-3x - 2 if if xs1 1<x< 6 if x26 below:
Evaluate the following integrals. Integral x*ln(x^2+3x+2)dx 2. (8 points) Evaluate the following integrals. Answers without supporting work will re- ceive no credit. (a) (5 points) | < ln(x2 + 3x + 2) dx
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
Evaluate piecewise-defined functions Question Given the following piecewise function, evaluate /(-4). - 4x + 3 f(x) = x < 0 Osr<3 3S 2? + 2 Do not include "f(-4) =" in your answer. Provide your answer below:
Let f (2) be defined by: k-?, <<-1 f(3) = z? +, -1<x<1 - kr1 Which of the following values of k would make f (2) continuous on R? Ok=0 There is no such value for k Ok= -1 Ok= 1