calculus problem...help!?

Question

Use linear approximation, i.e. the tangent line, to approximate
16.2^(1/2) as follow:

Let f(x)=x^(1/2). The equation of the tangent line to f(x) at x=16 can be written in the form of y=mx+b where m is:______. and where b is:______.

Using this, we find our approximation for 16.2^(1/2) is ______.

Answer 1

f(16) = 4
f'(16) = 1/8. Since this is the derivative, it gives you the slope of the tangent line at x=16.
You now have a point (4,16). Plug in to find b
So y=(1/8)x+b
(16) - (1/8)(4) = b, b = 2
So y=(1/8)x+2
And so approximation for 16.2^(1/2) is 4.025 because you would just plug in 16.2 in place of y in that last equation you found.

Answer 2

y' = 1/(2sqrt(x)) = 1/8 when x =16
m = 1/8
b = 2
y = x/8 + 2
When x = 16. 2 we have y = 4.025
Check:
4.025^2 = 16.200625