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

The normal line is given by

You can verify this on a graph: graph{(y-(2x^2+1)/x)(y+x+4)((y+3)^2+(x+1)^2-0.01)=0 [-10, 10, -5, 5]}

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To find the equation of the line normal to the function f(x) = (2x^2 + 1) / x at x = -1, we need to determine the slope of the tangent line at x = -1 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 the function f(x) with respect to x.

The derivative of f(x) = (2x^2 + 1) / x can be found using the quotient rule, which states that the derivative of (u/v) is (v * du/dx - u * dv/dx) / v^2.

Applying the quotient rule, we have:

f'(x) = [(x * d(2x^2 + 1)/dx) - (2x^2 + 1) * d(x)/dx] / x^2

Simplifying this expression, we get:

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

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

Now, to find the slope of the tangent line at x = -1, we substitute x = -1 into f'(x):

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

f'(-1) = -1

Therefore, the slope of the tangent line at x = -1 is -1.

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)

Slope of normal line = 1

Now, we have the slope of the normal line, and we also know that it passes through the point (-1, f(-1)). To find the y-coordinate of this point, we substitute x = -1 into the original function f(x):

f(-1) = (2(-1)^2 + 1) / -1

f(-1) = (2 + 1) / -1

f(-1) = -3

Therefore, the point (-1, f(-1)) is (-1, -3).

Using the slope-intercept form of a line, y = mx + b, where m is the slope and b is the y-intercept, we can substitute the values we have:

y = 1x + b

Since the line passes through (-1, -3), we can substitute these coordinates to find the value of b:

-3 = 1(-1) + b

-3 = -1 + b

b = -2

Therefore, the equation of the line normal to f(x) = (2x^2 + 1) / x at x = -1 is:

y = x - 2

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