Find the linear approximation of the function f(x) = √16-x at a = 0 and use it to approximate the numbers √15.9 and √15.99....
im trying to understand what to do and solving

Question

anonymous · Answer

To approximate a value of a function at a nasty point (for instance, $$\sqrt{15.99}$$), we can use a nicer value (like 16) and apply an "error" factor.  The formula for calculating the linear approximation is given by $$y=f(a) + f \prime(a)(x-a)$$
where a is the nice value and x is the nasty value.  So, in your case, you have $$f(x) = \sqrt{16-x}, a=0$$
$$f\prime(x)=-1/(2 \sqrt{16-x})$$
So, $$y=f(a) + f \prime(a)(x-a)$$
becomes
So, $$y=\sqrt{16-a}-[1/(2 \sqrt{16-a})](x-a)$$
Substituting a=0, we get
$$y=\sqrt{16}-[1/(2 \sqrt{16})](x)$$
$$=4-(x/8).$$
So the two values you are looking for correspond to x=.1 and x=.01 (since the original function was 16-x in the radical).
Using these values, we get approximations of 3.9875 and3.99875 respectively.

Hope that helps.