First, let's draw the graph and see what the question means.

y = 2x³ − x + 3

I have used equal scaling along the 2 axes (so that later, when I draw the circle, it will not have an elliptical shape).

Now, to find the radius of curvature, we need:

And then

So now we are ready to substitute into the formula to give us the radius at any point x:

Now to find the radius of curvature at the required point x = 1, we substitute:

To show what we have done, let's look at the graph of the curve (blue) with the approximating circle (black) overlaid. The circle is a good approximation for the curve at (1, 1).

We can show that the centre of the approximating circle is (-9.8, 6.17).