Limits, Derivative, Slope of Tangent and Equation of Tangent of a function

Question

Consider the function f(x)= 1÷ x+3

1. Use the limit definition of the derivative to find the derivative of the function.
2. Find the slope of the tangent to f(x) at x=0
3. Find the equation of the tangent to f(x) at x=0

Answer 1

f(x) = 1/(x+3) has derivative f'(x) = -1/(x+3)^2. at x= 0 the derivative = -1/9

the tangent line is y=-x/9 + b. find b by plugging in the point (0, 1/3)

b= 1/3, the tangent line is y=-x/9+1/3 or 9y= x + 3

find the derivative with the definition of a limit as h approaches zero for (f(x+h) - f(x))/h

= ( 1/(x+h+3) - 1/(x+h) )/h

= ( (x+3) - (x+h+3) )/(x+h+3)(x+h) )/ h

= -1/(x+3)^2

this all assumes you really meant f(x) = 1/(x+3) and not (1/x) +3. the way it's written you really have (1/x) +3, but that has no derivative at x=0. It's undefined.