What is the equation of the tangent line of #f(x)=sqrt(x^2-4x+7)/(x-1)# at #x=3#?

Question

Isabella Ackerman · Accepted Answer

#y = -1/4x + 7/4#

Start by finding the y-coordinate.

#f(3) = sqrt(3^2 - 4(3) + 7)/(3- 1) = sqrt(4)/2 = 1#

Differentiate using the quotient rule. Let #f(x) = (g(x))/(h(x))#, with #g(x) = sqrt(x^2 - 4x + 7)# and # h(x) = x - 1#.

We use the chain rule to differentiate #g(x)#:

Let #y = u^(1/2)# and #u = x^2 - 4x + 7#. Then #dy/(du) = 1/(2u^(1/2))# and #(du)/dx = 2x - 4#.

#g'(x)= dy/(du) * (du)/dx#

#g'(x) = 1/(2u^(1/2)) * 2x - 4#

#g'(x) = (2x- 4)/(2sqrt(x^2 - 4x + 7))#

#g'(x) = (2(x - 2))/(2sqrt(x^2 - 4x + 7))#

#g'(x)= (x- 2)/sqrt(x^2 - 4x + 7)#

Now use the quotient rule:

#f'(x) = (g'(x) * h(x) - g(x) * h'(x))/(h(x))^2#

#f'(x) = ((x - 2)/sqrt(x^2 - 4x + 7)(x - 1) - sqrt(x^2 - 4x + 7)(1))/(x - 1)^2#

#f'(x) = ((x^2 - 2x - x + 2)/sqrt(x^2 - 4x + 7) - sqrt(x^2 - 4x + 7))/(x- 1)^2#

#f'(x) = ((x^2 - 3x + 2 - (x^2 - 4x + 7))/sqrt(x^2 - 4x + 7))/(x - 1)^2#

#f'(x) = (x - 5)/(sqrt(x^2 - 4x + 7)(x- 1)^2#

The slope of the tangent line can be obtained by evaluating the point #x = a# into the derivative.

#f'(3) = (3 - 5)/(sqrt(3^2 - 4(3) + 7)(3 - 1)^2#

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

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

The equation of the tangent is therefore:

#y - y_1 = m(x - x_1)#

#y- 1 = -1/4(x - 3)#

#y - 1 = -1/4x + 3/4#

#y = -1/4x + 7/4#

Hopefully this helps!

Isabella Ackerman · Answer

undefined To find the equation of the tangent line of f(x) = sqrt(x^2 - 4x + 7)/(x - 1) at x = 3, we need to find the derivative of f(x) and evaluate it at x = 3. The derivative of f(x) can be found using the quotient rule:

f'(x) = [(x - 1)(2x - 4) - (x^2 - 4x + 7)(1)] / (x - 1)^2

Simplifying this expression gives:

f'(x) = (2x^2 - 6x + 4 - x^2 + 4x - 7) / (x - 1)^2
       = (x^2 - 2x - 3) / (x - 1)^2

Now, we can evaluate f'(x) at x = 3:

f'(3) = (3^2 - 2(3) - 3) / (3 - 1)^2
      = (9 - 6 - 3) / 2^2
      = 0

The slope of the tangent line at x = 3 is 0. To find the equation of the tangent line, we need a point on the line. We can use the point (3, f(3)).

f(3) = sqrt(3^2 - 4(3) + 7)/(3 - 1)
     = sqrt(9 - 12 + 7)/2
     = sqrt(4)/2
     = 2/2
     = 1

Therefore, the point on the tangent line is (3, 1). The equation of the tangent line can be written in point-slope form:

y - y1 = m(x - x1)

Using the point (3, 1) and the slope m = 0, we have:

y - 1 = 0(x - 3)
y - 1 = 0
y = 1

Hence, the equation of the tangent line of f(x) = sqrt(x^2 - 4x + 7)/(x - 1) at x = 3 is y = 1.