need help plz
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