Mathematics, 16.10.2019 01:30 ashlee7877
Find the poin on the graph of f(x)=1-x^2 that are closest to0(0,0)
Answers: 2
![\bf l=\left[ \boxed{x^2}~+~\left( \boxed{1-x^2} \right)^2\right]^{\frac{1}{2}}\implies l=(x^2+1-2x^2+x^4)^{\frac{1}{2}} \\\\\\ l(x)=(x^4-x^2+1)^{\frac{1}{2}}\\\\ -------------------------------\\\\ \cfrac{dl}{dx}=\cfrac{1}{2}(x^4-x^2+1)^{-\frac{1}{2}}(4x^3-2x)\implies \cfrac{dl}{dx}=\cfrac{4x^3-2x}{2\sqrt{x^4-x^2+1}} \\\\\\ \cfrac{dl}{dx}=\cfrac{2x^3-x}{\sqrt{x^4-x^2+1}}](/tpl/images/0323/5657/185e0.png)
are indeed minima, and so, what's "y" when "x" is that? well
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