suppose that f is differentiable on an open interval I containing x0.
f'(x0)cannot equal 0. Let y0=f(x0)
use the linear approximation to f at x0 to find an approximate
solution to the equation f(x)=y near y0

Hello fmunshi!
Thanks for the interesting question. First of all, we have to find the
linear approximation to f at x0; or, in other words, we have to find a
line that is tangent to f at x0. This will be the linear
approximation.

We know that such a line has the form:
l(x) = m*x + b

Where m is the slope of the line while b is the y-intercept. Our job
is to find m and b. Since f is differentiable in an open interval
containing x0, then it is differentiable at x0. Therefore f'(x0)
exists, and is the slope of f at x0. So the slope of the linear
approximation to f at x0 will have to be f'(x0). Thus we have found m:

m = f'(x0)

Finding b is easy. We know that the line has slope f'(x0). We also
know that it passes throught the point (x0,f(x0)) (it must "touch" f
at x0). Therefore, we have the following equation:

l(x) = m*x + b
f(x0) = m*x0 + b
f(x0) = f'(x0)*x0 + b
b = f(x0)-f'(x0)*x0

And, knowing the fact that f(x0)=y0

b=y0 - f'(x0)*x0

So, we now know exactly the equation for the linear approximation:

l(x) = f'(x0)*x + y0 - f'(x0)*x0
l(x) = f'(x0)*(x-x0) + y0

Now you have to use this approximation to find x such that f(x)=y near
y0. In order to do it, we simply replace y in l(x) and solve for x.
Using the approximation instead of the actual function f is what
limits us to finding f(x)=y only for y that are near y0. For y that
are "far" from y0, probably the linear approximation will also be
"far" from the true f(x), making it useless.

y = f'(x0)*(x-x0) + y0
y - y0 = f'(x0)*(x-x0)
x - x0 = (y - y0/f'(x0))

x = x0 + (y - y0/f'(x0))

That's the solution you're looking for. I hope this was clear enough.
If you have any doubts regarding my answer, please request a
clarification before rating it.


Best luck!
elmarto