The potential energy of a particle on the x-axis is given by U+ 8xe ^-xsquared/19 where x is 1 meter and U is in Joules. Please find the point on the x-axis for which the potential is a maximum or minimum. I've worked it three times and I get these three answers. Which one is correct?


4.359 m
3.082 m
1.202 m

How did you get those answers?

Where is the = sign in your equation?

To find the point on the x-axis for which the potential energy is either a maximum or a minimum, we need to take the derivative of the potential energy function with respect to x and set it equal to zero. This is because at a maximum or minimum point, the slope (or derivative) of the function is zero.

Let's first find the derivative of the potential energy function given by U + 8xe^(-x^2/19) with respect to x using the chain rule:

dU/dx = d(U)/dx + d(8xe^(-x^2/19))/dx

Now, let's calculate each term separately:

dU/dx = 0 (since U is a constant, it doesn't contribute to the derivative)

d(8xe^(-x^2/19))/dx = 8e^(-x^2/19) + 8x(-2/19)e^(-x^2/19)
= 8e^(-x^2/19) - (16x/19)e^(-x^2/19)
= (8 - (16x/19))e^(-x^2/19)

Setting this derivative equal to zero, we have:

(8 - (16x/19))e^(-x^2/19) = 0

To find the points where this equation is true, we need to solve for x:

8 - (16x/19) = 0

Solving this equation for x, we get:

(16x/19) = 8
16x = 8 * 19
x = (8 * 19) / 16
x = 76 / 16
x = 4.75

Therefore, the correct answer is x = 4.75 meters, not any of the options you provided.