Use linear approximation, i.e. the tangent line, to approximate square root (49.
ID: 3188961 • Letter: U
Question
Use linear approximation, i.e. the tangent line, to approximate square root (49.3) as follows:Let f(x) = square root (x). The equation of the tangent line to f(x) at x = 49 can be written in the form y = mx+b where m is: _________________ and where b is:_________________
Using this, we find our approximation for square root (49.3) is
NOTE: For this part, give your answer to at least 9 significant figures or use fractions to give the exact answer. _________________________
Explanation / Answer
We see that: f(x) = ?x = x^(1/2) f'(x) = (1/2)x^(-1/2) = 1/(2?x). Then, the slope of the tangent line at x = 49 is: f'(49) = 1/(2?49) = 1/(2 * 7) = 1/14. Next, f(x) passes through (49, 7). So our tangent line must to as well. With point-slope form, the equation of the line is: y - 7 = 1/14 * (x - 49) ==> y - 7 = (1/14)x - 7/2 ==> y = (1/14)x + 7/2. So: ?x ? (1/14)x + 7/2 when x ? 49. Thus: ?(49.3) ? (1/14)(49.3) + 7/2 ? 7.02143.
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