You use your calculus.
dU/dx, set to zero, solve for x
I don't know where the 19 goes, but take the derivative, set to zero
remember d/dx (uv)= vdu/dx + u dv/dx
The potential energy of a particle on the x-axis is given by U= 8xe ^-x ^2/19 where x is 1 meter and U is in Joules. Can you explain for me how to find the point on the x-axis for which the potential is a maximum or minimum and is this answer a point of maximum potential energy or minimum potential energy? Please show me how to figure this. Thanks
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