HELP PLEASE ???f ' (x) for a given function is (x2)/(x+3) and we are told that f (6) = 30. Write the equation of the tangent line to f(x) at x = 6 and use the tangent line equation to approximate the value of f (6.02).

Question

anonymous · Answer

29.92

30.02

30.08

34.00

none of these

anonymous · Answer

D

anonymous · Answer

$$f(x)=\frac{x^2}{x+3}\Rightarrow f'(x)=\frac{x^2+6x}{(x+3)^2}$$
The tangent line at x = 6 is given by
$$y-y_0=m(x-x_0),\text{ where }(x_0,y_0)=(6,30),\text{ and }m=f'(6).$$
You then use this equation to find an approximate value of f(6.02) by plugging in x = 6.02.