Consider the function

f(x)=C(2xx3)0<x<520otherwise

Could fbe a probability density function? If so, determine

C. Repeat if f(x)were given by

localid="1646632841793" f(x)=C(2xx3)0<x<520otherwise

Short Answer

Expert verified

Whenc=0,±2thenf(x)<0

Step by step solution

01

Given Information

f(x)=C(2xx3)0<x<520otherwise

02

Explanation

2x-x3=-x(x-2)(x+2)

c=0

-f(x)dx=0,-f(x)dx=1

c>0,f(x)<0,x(2,52)

03

Explanation

c=2

f(x)=2(2x-x3)=4x-2x3

f(x)<0,x(2,52)

f(1.5)=4×15-2×(1.5)3=-0.75

c<0,f(x)<0(0,2)

04

Explanation

c=-2

f(x)=-2(2x-x3)

role="math" localid="1646639321144" =-4x+2x3

f(1)=-4+(2×1)=-2

f(x)0

05

Explanation

f(x)=C(2xx3)0<x<520otherwise

2x-x2=-x(x-2)

c=0,-f(x)dx=0

c>0,f(x)<0,x(2,52)

06

Explanation

x=2.2

f(2.2)=4(2.2)-2(2.2)=-0.88

c<0,x(0,2)

c=-2,f(x)=-4x+2x2

f(1)=-4+2=-2<0

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