Determine the minimal distance from point (-3,3) to the curve given by y=(x-3)^2
Let [mathA(-3,3), \ P(x,(x-3)^2)[/tex]
$\displaystyle AP^2=(x+3)^2+[(x-3)^2-3]^2=x^4-12x^3+49x^2-66x+45=f(x)$
$\displaystyle f'(x)=2(2x^3-18x^2+49x-33)=2(x-1)(2x^2-16x+33)$
$\displaystyle f'(x)=0\Rightarrow x=1$
If $\displaystyle x<1\Rightarrow f'(x)<0\Rightarrow$ f decreasing.
If $\displaystyle x>1\Rightarrow f'(x)>0\Rightarrow$ f increasing.
Then $\displaystyle x=1$is a point of minimum. The minimum of f is $\displaystyle f(1)=17$. The minimum distance is $\displaystyle \sqrt{17}$