Find an equation of the tangent line to the graph of the function at the given point. f(x) = (x+1/x−3)^2 ; (7, 4)

Answer 1

As f(g(x))' = f'(g(x))g'(x) and (f(x)/g(x))' = (f'(x)g(x)-g'(x)f(x))/(g(x))^2, we get

(x+1/x-3)^2 ' = 2 (x+1/x-3) (1*(x-3)-1^(x+1))/(x-3)^2 = -8(x+1)/(x-3)^3

Thus, f'(7) = -8(7+1)/(7-3)^3 = -8(8)/4^3 = -64/64 = -1.

Then, as f(7) = 4, we have

y = mx+b, or

4 = (-1)7 + b

11 = b

The tangent line is y = -x + 11