# 查找$x \ sqrt {16-x ^ {2}}$的最大/最小值

## 最佳答案

I believe you have found the incorrect critical points. To find them, you set $f'(x)=0$, in this case, \begin{align*} \sqrt{16-x^{2}} -\frac{x^{2}}{\sqrt{16-x^{2}}}=0 &\implies \sqrt{16-x^2}=\frac{x^{2}}{\sqrt{16-x^{2}}} \\ &\implies 16-x^2=x^2 \\ &\implies x^2=8 \\ &\implies x=\pm 2\sqrt{2} \end{align*} So your critical points are $x=-4,\pm 2\sqrt{2},4$, when you include the endpoints of your interval. Plugging back in, you find $f(4)=f(-4)=0$, and $f(2\sqrt{2})=8$ and $f(-2\sqrt{2})=-8$. From this you can conclude what the relative extrema are on your interval.