Shouldn't be too hard for you

$x+\dfrac{1}{x}\geq 2$

So answer is $\boxed{20}$

By AM-GM, we get $10x+\frac{10}{x}\geq 2\sqrt{10x\cdot\frac{10}{x}}=2\sqrt{10\cdot10}=2\cdot10=20$

Taking the derivative and setting it to zero gives you

10 - 10x^-2 = 0

Then x^2 = 1, and x = +1 or -1 are the interesting points of this function.

At x = +1 you get 20. At x = -1 you get -20.

You can tell that the function increases to either side of x = +1 using more calculus or just trying some points (x = 1/10 gives 101, as does x = 10), so there's a local minimum at x = +1.

You can actually make the function arbitrarily small using negative values of x. x = -10 gives a result of -101. As x goes to negative infinity, the function goes to 10x. Because I can't type negative infinity into the answer box, I decided you meant to restrict to positive values of x. But I think you should edit the question to make this clear, if that is a thing you can do.

It's minimum when 10x = 10/x, Because x is positive we get x=1. Then the minimum value is 20