9514 1404 393
Answer:
0.0056
Step-by-step explanation:
f(x) = √(49 +x)
f'(x) = 1/(2√(49 +x))
A linear approximation of f(x) expanded about x=0 is ...
f(x) ≈ f(0) + f'(0)x = 7 +x/(2·7)
Then for √53, we have x=4
f(4) ≈ 7 +4/14 = 7 2/7 . . . . . approximate √53 using differentials
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The calculator value of √53 is about 7.280110, so the difference in results is ...
approx - actual ≈ 7.285714 -7.280110 = 0.005604 ≈ 0.0056
The lines have to be the same
