0 0 votes The point on $y=x^2+1$ closest to $(0,2)$ is $(0.707,1.5)$ $(0.707,-1.5)$ $(-0.707,1.5)$ $(-0.707,-1.5)$ Calculus goclasses calculus-&-optimization goclasses-da-dpp goclasses-da-dpp-day-54 goclasses-calculus-&-optimization-practice-questions + – GO Classes 238 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes Objective function $f(x)=(x-0)^2+\left(x^2+1-2\right)^2$ $$ f(x)=x^4-x^2+1 $$ For minima $f^{\prime}(x)=0$ $$ \begin{aligned} & 4 x^3-2 x=0 \\ & x=0,0.707,-0.707 \end{aligned} $$ Corresponding $y=1,1.5,1.5$ GO Classes answered Nov 25, 2025 • edited Nov 25, 2025 by GO Classes GO Classes comment Share Follow 0 reply Please log in or register to add a comment.