Hello, Harry!

Just take the derivative, equate to zero, and solve . . .

Assuming x > 0, we have: .f(x) .= .2ax·e^{-ax²}

Differentiate and equate to zero:

. . f'(x) .= .2ax·e^{-2ax²}(-2ax) + 2ae^{-ax²} .= .0

Divide by 2ae^{-ax²}: .-2ax² + 1 .= .0

And we have: .2ax² .= .1 . . → . . x² = 1/2a

. . . . . . . . . . . . . . .__

Therefore: .x .= .1/√2a