Independent solution

How to solve this Continuous Random Variables question

Setup

Setup

The positive constant multiplying the density does not affect its mode. Shift the variable two units to center the remaining polynomial, then maximize it on the stated support.

g(x)=3(x−2)2−(x−2)4+4g(x)=3(x-2)^2-(x-2)^4+4
g′(x)=6(x−2)−4(x−2)3g'(x)=6(x-2)-4(x-2)^3

Model

Model

Factor the derivative and retain only stationary points inside the support.

g′(x)=2(x−2)(3−2(x−2)2)=0g'(x)=2(x-2)\left(3-2(x-2)^2\right)=0
x=2,x=2±32x=2,\qquad x=2\pm\sqrt{\frac32}
2+32>32+\sqrt{\frac32}>3

Compute

Compute

Compare the admissible stationary points with the boundary limits. The lower critical point gives the largest density height.

lim⁡x↓0f(x)=0,f(2)=518=0.277778\lim_{x\downarrow0}f(x)=0,\qquad f(2)=\frac5{18}=0.277778
f(2−32)=125288=0.434028f\left(2-\sqrt{\frac32}\right)=\frac{125}{288}=0.434028
lim⁡x↑3f(x)=512=0.416667\lim_{x\uparrow3}f(x)=\frac5{12}=0.416667

Answer

Answer

The density reaches its unique maximum at approximately 0.775.

2−32=0.775255…(B)\boxed{2-\sqrt{\frac32}=0.775255\ldots\quad\text{(B)}}