153k views
0 votes
Find the absolute extrema of the function. f(x) = xe^(-x^2) on [0 2]

User Ragamufin
by
8.7k points

1 Answer

6 votes

f(x)=xe^(-x^2)

f'(x)=(1-2x^2)e^(-x^2)

f''(x)=(4x^3-6x)e^(-x^2)


f'(x)=0 whenever
1-2x^2=0, which happens when
x=\pm\frac1{\sqrt2}. Only the positive root falls in the interval
[0,2]. At this point, you have
f''\left(\frac1{\sqrt2}\right)<0 (concave downward). This indicates a maximum at
\left(\frac1{\sqrt2},\frac1{√(2e)}\right)\approx0.7071,0.4289).

Meanwhile,
f(0)=0 and
f(2)=2e^(-4)\approx0.0366, so the maximum above must be absolute, and the absolute minimum occurs at the origin.
User Mike Patrick
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories