How do you find the equation of the line tangent to #f(x)= 2/(x-1)# at x=3?

Question

Evelyn Agard · Accepted Answer

#y=-1/2x+5/2#

To find the equation of the line tangent to the given function, you will need to:

#1#. Find the derivative.
#2#. Substitute #x=3# into the given function to find the y-coordinate
#3#. Substitute #x=3# into the derivative to find the slope of the tangent.
#4#. With the acquired data, use the slope formula to determine the equation of the tangent.
Finding The Derivative
In your case, to find the derivative, you will need to use four rules - the constant multiple rule, the reciprocal rule, the difference rule, and the power rule. The rules can be expressed as follows:

#color(red)("Constant Multiple Rule")#: #[cf(x)]'=c[f(x)]'#
#color(green)("Reciprocal Rule")#: #[1/f(x)]'=-(f'(x))/[f(x)]^2#
#color(blue)("Difference Rule")#: #[f(x)-g(x)]'=f'(x)-g'(x)#
#color(purple)("Power Rule")#: #(x^n)'=nx^(n-1)#

#1#. Start by using the #color(red)("constant multiple rule")# to factor the numerator, #2#, from the function since it is a constant .

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

#2#. With the #color(green)("reciprocal rule")#, apply the necessary changes as stated in its formula.

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

#3#. Using the #color(blue)("difference rule")# first and then the #color(purple)("power rule")#, apply the necessary changes as stated in their formulas.

#f'(x)=-2[((1-0))/(x-1)^2]#
#f'(x)=-2[1/(x-1)^2]#
#f'(x)=-2/(x-1)^2#

Finding the Coordinates of the Point
To find the y-coordinate of the point when #x=3#, substitute #x=3# into the function.

#f(x)=2/(x-1)#
#f(3)=2/(3-1)#
#f(3)=1#

Finding the Slope of the Tangent
To find the slope of the equation tangent to the point #(3,1)#, substitute #x=3# into the derivative.

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

Finding the Equation of the Tangent
Using the equation of a line in point-slope form, substitute your known values into the equation. Then rewrite the equation in slope-intercept form:

#y-y_1=m(x-x_1)#
#y-1=-1/2(x-3)#
#y-1=-1/2x+3/2#
#y=-1/2x+5/2#

If you graph the given function and the tangent, you will see that the line intersects #(3,1)# at one point:

#:.#, the equation of the line tangent to the given function is #y=-1/2x+5/2#.

Axel Alvarez · Answer

undefined To find the equation of the line tangent to f(x) = 2/(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) = [2 * (x - 1)^0 - 2 * 1 * (x - 1)^(-1)] / (x - 1)^2

Simplifying this expression gives:

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

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

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

So, the slope of the tangent line at x = 3 is -1/2. We also know that the point (3, f(3)) lies on the tangent line. Plugging x = 3 into f(x), we get:

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

Therefore, the point (3, 1) lies on the tangent line. Using the point-slope form of a line, we can write the equation of the tangent line as:

y - 1 = (-1/2)(x - 3)

Simplifying this equation gives:

y - 1 = (-1/2)x + 3/2

Finally, rearranging the equation to the slope-intercept form, we have:

y = (-1/2)x + 3/2 + 1
y = (-1/2)x + 5/2

Hence, the equation of the line tangent to f(x) = 2/(x-1) at x = 3 is y = (-1/2)x + 5/2.