# What is the equation of the normal line of #f(x)=(x+5)/(e^x-1)# at #x=1#?

Normal Line:

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To find the equation of the normal line of f(x)=(x+5)/(e^x-1) at x=1, we need to find the slope of the tangent line at x=1 and then determine the negative reciprocal of that slope to find the slope of the normal line.

To find the slope of the tangent line, we can take the derivative of f(x) with respect to x and evaluate it at x=1.

The derivative of f(x) is given by f'(x) = [(e^x-1)(1) - (x+5)(e^x)] / (e^x-1)^2.

Evaluating f'(x) at x=1, we get f'(1) = [(e^1-1)(1) - (1+5)(e^1)] / (e^1-1)^2.

Simplifying this expression, we find f'(1) = (-6e + e - 1) / (e - 1)^2.

Now, to find the slope of the normal line, we take the negative reciprocal of f'(1).

The slope of the normal line is -1 / f'(1).

Finally, we can use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point of tangency (in this case, x=1, f(1)), and m is the slope of the normal line.

Plugging in the values, the equation of the normal line is y - f(1) = (-1 / f'(1))(x - 1).

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