Final answer:
The equation of the normal line to the curve y=2xe^x at the point x=1 is y = -1/(4e)(x - 1) + 2e.
Step-by-step explanation:
To find the equation of the normal line to the curve y=2xe^x at the point x=1, we first need to find the derivative of the curve to determine the slope of the tangent line at that point. The derivative of y with respect to x, denoted as y', is found by applying the product rule:
y' = 2e^x + 2xe^x. At x=1, the derivative y' evaluates to y'(1) = 2e + 2e = 4e, which is the slope of the tangent line.
Since the normal line is perpendicular to the tangent line, its slope is the negative reciprocal of the tangent line's slope, which is -1/(4e).
Now we can use the point-slope form of a line to find the equation of the normal line:
y - y1 = m(x - x1),
where (x1, y1) is the point on the curve and m is the slope of the normal line. Substituting the values, we get
y - 2e = -1/(4e)(x - 1).
The final equation of the normal line at x=1 is y = -1/(4e)(x - 1) + 2e.