# What is the equation of the line that is normal to #f(x)= x/sqrt( 3x+2) # at # x=4 #?

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To find the equation of the line that is normal to the function f(x) = x/√(3x+2) at x=4, we need to determine the slope of the tangent line at x=4 and then find the negative reciprocal of that slope to obtain the slope of the normal line.

First, we find the derivative of f(x) with respect to x, which is given by:

f'(x) = (3x - 2)/(2√(3x+2))

Next, we substitute x=4 into the derivative to find the slope of the tangent line at x=4:

f'(4) = (3(4) - 2)/(2√(3(4)+2)) = (12 - 2)/(2√(14)) = 10/(2√(14)) = 5/√(14)

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

m_normal = -1/(5/√(14)) = -√(14)/5

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 line passes through the point (4, f(4)), so we substitute x=4 into the original function to find the corresponding y-coordinate:

f(4) = 4/√(3(4)+2) = 4/√(12+2) = 4/√14

Using the point-slope form, the equation of the line that is normal to f(x) at x=4 is:

y - (4/√14) = (-√14/5)(x - 4)

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