 # How do you find the point on the curve y=2x^2 closest to (2,1)? Lokubovumn 2022-08-13 Answered
How do you find the point on the curve $y=2{x}^{2}$ closest to (2,1)?
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Every point on the curve has the form: $\left(x,2{x}^{2}\right)$
The closest point is the one whose distance is minimum.
The distance between $\left(x,2{x}^{2}\right)$ and (2,1) is $\sqrt{{\left(x-2\right)}^{2}+{\left(2{x}^{2}-1\right)}^{2}}$
Because the square root is an increasing function, we can minimize the distance by minimizing:
$f\left(x\right)={\left(x-2\right)}^{2}+{\left(2{x}^{2}-1\right)}^{2}$
$f\left(x\right)=4{x}^{4}-3{x}^{2}-4x+5$
To minimize f
$f\prime \left(x\right)=16{x}^{3}-6x-4$
Now use whatever tools you have available to solve this cubic equation. (Formula, graphing technology, successive approximation, whatever you have.)
The critical number is approximately 0.824.
Clearly f'(0) is negative and f'(1) is positive, so f(0.824) is a local minimum.
f is a polynomial with only one critical number, so "local" implies "global"
The minimum value of distance occurs at $x\approx 0.824$ and $f\left(0.824\right)\approx 1.358$
The closest point is about (0.824,1.358).