the equation of the tangent line to f(x)=sqrt(x) at x=16
y=
b=
using this, find approximation for sqrt(16.2)
sooo lost =(
Okay not to worry you, but maybe the most important point from Calc I is the fact that the derivative of a function is the slope of the function... so the derivative at a point x, is the slope of the function at the point x
A tangent line has the same slope as the curve at the point of intersection, thus it's slope is equal to the derivative
If you don't remember anything else from Calc I, remember that
You plugged 16 into f(x) to get y, so y=4, and m is the derivative, which you have there, evaluated at x=16
So $\displaystyle f'(x)=\frac{1}{2\sqrt{x}}$
And $\displaystyle m=f'(16)=\frac{1}{2\sqrt{16}}$