Are there any points on the curve y = x - 1/(2x) where the slope is 2? If so, find them.

Verified step by step guidance

1. To find points on the curve where the slope is 2, we need to first determine the derivative of the function y = x - \frac{1}{2x}. The derivative, y', represents the slope of the tangent line at any point on the curve.
2. Differentiate the function y = x - \frac{1}{2x} with respect to x. The derivative of x is 1, and the derivative of -\frac{1}{2x} can be found using the power rule. Rewrite -\frac{1}{2x} as -\frac{1}{2}x^{-1} and differentiate to get \frac{1}{2}x^{-2}.
3. Combine the derivatives: y' = 1 + \frac{1}{2}x^{-2}. Simplify this to y' = 1 + \frac{1}{2x^2}.
4. Set the derivative equal to 2, since we are looking for points where the slope is 2: 1 + \frac{1}{2x^2} = 2.
5. Solve the equation 1 + \frac{1}{2x^2} = 2 for x. Subtract 1 from both sides to get \frac{1}{2x^2} = 1. Multiply both sides by 2x^2 to isolate x^2, resulting in 1 = 2x^2. Solve for x by dividing both sides by 2 and taking the square root, giving x = \pm\frac{1}{\sqrt{2}}. Substitute these x-values back into the original equation y = x - \frac{1}{2x} to find the corresponding y-values.