FIND THE APPROXIMATE VALUE USING LINEAR APPROXIMATION

Let f : (a,b)→R be a differentiable function and

x∈ (a, b)

We define the linear approximation L of f at x0 by

Problem 1 :

Let f (x)  =  x . Find the linear approximation at x = 27. Use the linear approximation to approximate ∛27.2

L(x)  =  f(x0) + f'(x0) (x-x0)

Solution :

f(x)  =  x

Let x0  =  27 and x  =  27.2

L(x)  =  f(x0) + f'(x0) (x - x0)

f(x)  =  x

f(x0)  =  ∛27

f(x0)  =  3

f'(x)  =  (1/3)x-2/3

f'(x0)  =  (1/3)(27)-2/3

f'(x0)  =  (1/3)(1/9)

f'(x0)  =  1/27

L(27.2)  =  3 + (1/27) (27.2-27)

L(27.2)  =  3 + (1/27) (0.2)

L(27.2)  =  3 + 0.0074

L(27.2)  =  3.0074 (approximately)

Problem 2 :

Use the linear approximation to find approximate values of

(i)  (123)2/3

Solution :

f(x)  =  (x)2/3

Let x  =  123 and x0  =  125 (the nearest whole number)

f(x)  =  (x)2/3

f(125)  =  (125)2/3

f(125)  =  (53)2/3

 f(125)  =  25



f'(x)  =  (2/3)x2/3 - 1

f'(x)  =  (2/3)x-1/3

f'(125)  =  (2/3)(125)-1/3

f'(125)  =  (2/3)(53)-1/3

f'(125)  =  (2/3)(5-1)

f'(125)  =  2/15

L(x)  =  f(x0) + f'(x0) (x - x0)

L(123)  =  25 + (2/15) (123-125)

L(123)  =  25 + (2/15) (-2)

L(123)  =  25 - (4/15)

L(123)  =  25 - 0.2667

L(123)  =  24.733

So, the approximate value of (123)2/3 is 24.733.

(ii)  (15)1/4

Solution :

L(x)  =  f(x0) + f'(x0)(x-x0)

Let x0  =  16 and x  =  15

L(15)  =  f(16) + f'(16)(x-x0)  ----(1)

f(x)  =  x1/4

f(16)  =  161/4

f(16)  =  (24)1/4

f(16)  =  2


f'(x)  =  (1/4) x1/4 - 1

f'(x)  =  (1/4) x-3/4

f'(16)  =  (1/4) (16)-3/4

f'(16)  =  (1/4) (2-3)

f'(16)  =  (1/32)

L(15)  =  f(16) + f'(16)(x-x0)  ----(1)

By applying the values in (1), we get

L(15)  =  2 + (1/32)(15-16)

L(15)  =  2 + (1/32)(-1)

L(15)  =  2 - 0.03125

L(15)  =  1.968 (approximately)

So, the approximate value of (15)1/4 is  1.968.

(iii)  ∛26

Solution :

Let f(x)  =   ∛x

Let x  =  26 and x0  =  27

L(x)  =  f(x0) + f'(x0) (x-x0)

L(26)  =  f(27) + f'(27) (x-x0)  ----(1)

f(x)  =   ∛x

f(27)  =  ∛27

f(27)  =  3

f'(x)  =  (1/3) x1/3-1

f'(x)  =  (1/3) x-2/3

f'(27)  =  (1/3) 27-2/3

f'(27)  =  (1/3) (1/9)

f'(27)  =  1/27

By applying these value in (1), we get

L(26)  =  3 + (1/27) (27-26)

L(26)  =  3 + (1/27) (1)

L(26)  =  3 + 0.0370

L(26)  =  3.0370

So, the approximate value of ∛26 is 3.0370.

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