# What is the equation of the normal line of #f(x)= (1+2x)*e^(x^2)# at #x=-1/2#?

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To find the equation of the normal line of f(x) = (1+2x)e^(x^2) at x = -1/2, we need to determine the slope of the tangent line at that point and then find the negative reciprocal of that slope to obtain the slope of the normal line.

To find the slope of the tangent line, we can use the derivative of f(x). Taking the derivative of f(x) with respect to x, we get:

f'(x) = (2 + 4x + 4x^3)e^(x^2)

Evaluating f'(-1/2), we substitute x = -1/2 into the derivative:

f'(-1/2) = (2 + 4(-1/2) + 4(-1/2)^3)e^((-1/2)^2)

Simplifying this expression, we find:

f'(-1/2) = (2 - 2 + 4/8)e^(1/4)

f'(-1/2) = (1/2)e^(1/4)

Therefore, the slope of the tangent line at x = -1/2 is 1/2e^(1/4).

To find the slope of the normal line, we take the negative reciprocal of the slope of the tangent line:

Slope of normal line = -1/(1/2e^(1/4))

Simplifying this expression, we get:

Slope of normal line = -2e^(-1/4)

Now that we have the slope of the normal line, we can use the point-slope form of a line to find the equation. We know that the point of tangency is (-1/2, f(-1/2)). Substituting these values into the point-slope form, we have:

y - f(-1/2) = -2e^(-1/4)(x - (-1/2))

Simplifying further, we obtain the equation of the normal line:

y - f(-1/2) = -2e^(-1/4)(x + 1/2)

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