1. equation of tangent line

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 =(

2. Originally Posted by sjara
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 =(

Can you find the equation of the tangent line? If you can, then plug x=16.2 into the equation of the tangenet line

3. this is what i've got so far..
f '(x)=1/2x^-1/2
so i plugged in 16 to get y

so now its y-4=m(x-16)
but i dont know how to get m

4. Originally Posted by sjara
this is what i've got so far..
f '(x)=1/2x^-1/2
so i plugged in 16 to get y

so now its y-4=m(x-16)
but i dont know how to get m

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 $f'(x)=\frac{1}{2\sqrt{x}}$

And $m=f'(16)=\frac{1}{2\sqrt{16}}$

5. i've been reading this book for hours and didnt understand the wording until you explained it the way you did. thanks again