# What is the equation of the normal line of #f(x)=x/sqrt(x+1/x)# at #x=2 #?

The slope of the normal is -1 / f'=--16.58, nearly.

So, the equation to the normal at P(2, 1.265) is

graph{(y-x/sqrt(x+1/x))(y+16.6x-34.4)((x-2)^2+(y-1.3)^2-.01)=0 [-10, 10, -5, 5]}

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

First, let's find the derivative of f(x) using the quotient rule:

f'(x) = [sqrt(x+1/x) - x(1/2)(x+1/x)^(-1/2)(1-1/x^2)] / (x+1/x)

Next, we substitute x=2 into f'(x) to find the slope of the tangent line at x=2:

f'(2) = [sqrt(2+1/2) - 2(1/2)(2+1/2)^(-1/2)(1-1/2^2)] / (2+1/2)

Simplifying this expression, we get:

f'(2) = [sqrt(5/2) - 2(1/2)(2+1/2)^(-1/2)(3/4)] / (5/2)

Now, we can find the negative reciprocal of f'(2) to obtain the slope of the normal line:

m_normal = -1 / f'(2)

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=2, f(2)), to find the equation of the normal line.

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