Question

Consider the function

$f\left(x\right)=\left\{\begin{array}{ll}C(2x-x^3)& 0<x<\frac{5}{2}\\ 0& otherwise\end{array}\right.$

Could $f$be a probability density function? If so, determine

$C$. Repeat if $f\left(x\right)$were given by

localid="1646632841793" $f\left(x\right)=\left\{\begin{array}{ll}C(2x-x3)& 0<x<\frac{5}{2}\\ 0& otherwise\end{array}\right.$

Short answer

When$c=0,\pm 2thenf\left(x\right)<0$

Step-by-step solution

Given Information

$f\left(x\right)=\left\{\begin{array}{ll}C(2x-x^3)& 0<x<\frac{5}{2}\\ 0& otherwise\end{array}\right.$

Explanation

$2x-x^3=-x(x-\sqrt{2})(x+\sqrt{2})$

$c=0$

${\int }_{-\infty }^{\infty }f\left(x\right)dx=0,{\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

$c>0,f\left(x\right)<0,x\in (\sqrt{2},\frac{5}{2})$

Explanation

$c=2$

$f\left(x\right)=2(2x-x^3)=4x-2x^3$

$f\left(x\right)<0,x\in (\sqrt{2},\frac{5}{2})$

$f(1.5)=4\times 15-2\times {(1.5)}^3=-0.75$

$c<0,f\left(x\right)<0(0,\sqrt{2})$

Explanation

$c=-2$

$f\left(x\right)=-2(2x-x^3)$

role="math" localid="1646639321144" $=-4x+2x^3$

$f\left(1\right)=-4+(2\times 1)=-2$

$f\left(x\right)\ge 0$

Explanation

$f\left(x\right)=\left\{\begin{array}{ll}C(2x-x^3)& 0<x<\frac{5}{2}\\ 0& otherwise\end{array}\right.$

$2x-x^2=-x(x-2)$

$c=0,{\int }_{-\infty }^{\infty }f\left(x\right)dx=0$

$c>0,f\left(x\right)<0,x\in (2,\frac{5}{2})$

Explanation

$x=2.2$

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

$c<0,x\in (0,2)$

$c=-2,f\left(x\right)=-4x+2x^2$

$f\left(1\right)=-4+2=-2<0$