Let
. Then differentiating, we get

We approximate
at
with the tangent line,

The
-intercept for this approximation will be our next approximation for the root,

Repeat this process. Approximate
at
.

Then

Once more. Approximate
at
.

Then

Compare this to the actual root of
, which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.
1-1, 1-2, 1-3, 1-4 and u should Change it to middle school
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