Question

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

Answer 1

Compute f'(x) and then evaluate f'(3) this is the slope then use the point slope equation of a line for the point (3, f(3))

Explanation:

Term-by-term
The derivative of the first term written as (x - 4)^-3 is

-3(x - 4)^-4

The derivative of the second term requires the product rule, (g(x)h(x))' = g'(x)h(x) + g(x)h'(x) or the quotient rule. I prefer the former:

let x = g(x), then g'(x) = 1, h(x) = (x - 2)^-1, and h'(x) = -(x - 2)^-2

Substituting into the product rule:

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

Collecting terms:

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

The slope, m, is f'(3):

m = f'(3) = -3(3 - 4)^4 + (3 - 2)^-1 - 3(3 - 2)^-2)

m = -3 + 1 - 3

m = -5

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

Using the point slope form y - y1 = m(x - x1) at the point (3,2):

y - 2 = -5(x - 3)