Local Linear Approximation; Differentials Solutions To Selected ...

3. Use an appropriate local linear approximation to estimate the value of the given
quantity.
(a) (3.02) 4
By inspection we see that can be f(x) = (x) 4 . Then x0 = 3 and △x =
0.02, with f ′ (x) = 4x 3 .
f(3 + △x) ≈ f(3) + f ′ (3) △ x
f(3.02) ≈ (3) 4 + 4(3) 3 (0.02)
= 81 + 108 2
100
= 81 + 108 1
50
= 81 + 54
25
= 81 + 2 4
25
= 83 4
= 83.16
25
(3.02) 4 ≈ 83.16 with the actual value to 4 decimals places of 83.181696.
(b) √ 65
The closest perfect square to this is √ 64 = 8. So f(x) = √ x,
f ′ (x) = 1
2 √ x , x0 = 64 and △x = 1.
f(64 + △x) = f(64) + f ′ (64) △ x
√ 65 ≈ √ 64 + 1
2 √ 64 (1)
= 8 + 1
16
= 8 1
= 8.0625
16
√ 65 ≈ 8.0625 with the actual value to 5 decimal places of 8.06225.
3