## anonymous 3 years ago 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).

1. anonymous

29.92 30.02 30.08 34.00 none of these

2. anonymous

D

3. anonymous

$f(x)=\frac{x^2}{x+3}\Rightarrow f'(x)=\frac{x^2+6x}{(x+3)^2}$ 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.

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