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

Question

Emilia Aguilar · Accepted Answer

The equation is:

#y(x) =-4/11x+71/22#

The general formula of the normal line to a curve is:

#y(x) = f(bar x) -1/(f'(bar x)) (x - bar x)#

Given:

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

For #x=2#:

#f(x)|_(x=2) =5/2#

#f'(x)|_(x=2) =11/4#

So:

#y(x) =5/2-4/11(x-2)=-4/11x+71/22#

Axel Alvarez · Answer

undefined To find the equation of the normal line of f(x) = x^2 - x + 1/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, we find the derivative of f(x) using the quotient rule:

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

Simplifying this expression, we get:

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

Combining like terms, we have:

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

Now, we can find the slope of the tangent line at x = 2 by substituting x = 2 into f'(x):

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

Simplifying, we get:

f'(2) = (4 - 1) / 4 = 3/4

The slope of the tangent line at x = 2 is 3/4.

To find the slope of the normal line, we take the negative reciprocal of 3/4:

m_normal = -4/3

Now, we have the slope of the normal line. To find the equation of the normal line, we use the point-slope form of a line and substitute the point (2, f(2)) = (2, 2^2 - 2 + 1/2) = (2, 7/2):

y - y1 = m(x - x1)

y - 7/2 = -4/3(x - 2)

Multiplying through by 6 to eliminate fractions, we get:

3y - 21 = -8(x - 2)

Simplifying further:

3y - 21 = -8x + 16

Rearranging the equation to the standard form:

8x + 3y = 37

Therefore, the equation of the normal line of f(x) = x^2 - x + 1/x at x = 2 is 8x + 3y = 37.