# What is the equation of the normal line of #f(x)=x/(x-1) # at #x=4 #?

derivative for slope , use 2 point form of a line

hence slope of normal is

by 2 point form

hope u find it helpful :)

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To find the equation of the normal line of f(x)=x/(x-1) 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.

To find the slope of the tangent line, we can take the derivative of f(x) with respect to x.

The derivative of f(x) = x/(x-1) can be found using the quotient rule, which states that the derivative of (u/v) is (v*u' - u*v') / v^2, where u' and v' represent the derivatives of u and v, respectively.

Applying the quotient rule, we have:

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

Simplifying this expression, we get:

f'(x) = -1 / (x-1)^2

Now, to find the slope of the tangent line at x=4, we substitute x=4 into f'(x):

f'(4) = -1 / (4-1)^2 = -1 / 9

The slope of the tangent line at x=4 is -1/9.

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

Slope of the normal line = -1 / (-1/9) = 9

Now that we have the slope of the normal line, we can use the point-slope form of a line to find its equation. We know that the normal line passes through the point (4, f(4)).

Substituting x=4 into f(x), we get:

f(4) = 4 / (4-1) = 4/3

Therefore, the point of intersection is (4, 4/3).

Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope, we can substitute the values:

y - (4/3) = 9(x - 4)

Simplifying this equation, we get:

y - 4/3 = 9x - 36

Rearranging the equation to the standard form, we have:

9x - y = 36 + 4/3 9x - y = 108/3 + 4/3 9x - y = 112/3

Therefore, the equation of the normal line of f(x)=x/(x-1) at x=4 is 9x - y = 112/3.

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