FINDING LINEAR APPROXIMATION FOR A FUNCTION

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

x∈ (a, b)

We define the linear approximation L of f at x0 by Find a linear approximation for the following functions at the indicated points.

Problem 1 :

f(x)  =  x3-5x+12 and x0  =  2

Solution :

f(x)  =  x3-5x+12

Let x0  =  2

L(x)  =  f(x0) + f'(x0) (x - x0)

 f(x)  =  x3-5x+12f(x0)  =  23-5(2)+12f(2)  =  8-10+12f(2)  =  10 f(x)  =  x3-5x+12f'(x)  =  3x2-5f'(x0)  =  3x2-5f'(2)  =  3(2)2-5f'(2)  =  12-5f'(2)  =  7

L(x)  =  f(2) + f'(2) (x - 2)  ----(1)

Applying the above values in (1), we get

L(2)  =  10 + 7(x - 2)

L(2)  =  10 + 7x - 14

L(2)  =  7x - 4

Problem 2 :

g(x)  =  √(x2+9) and x0  =  -4

Solution :

g(x)  =  √(x2+9) and x0  =  -4

Here f(x) and g(x) are equal.

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

g(x0)  =   √((-4)2+9)  =  √(42+9)  =  5

g'(x)  =  [1/2√(x2+9)] (2x)

g'(x)  =  x/√(x2+9)

g'(x0)  =  g'(-4)  =  -4/√(42+9)

g'(-4)  =  -4/5

By applying the values of g(x0) and g'(x0) in (1), we get

L(-4)  =  5 + (-4/5)(x+4)

L(-4)  =  5 - (4x/5) - (16/5)

L(-4)  =  (9/5) - (4x/5)

L(-4)  =  (1/5) (9-4x)

Problem 3 :

h(x)  =  x/(x+1) and x0  =  1

Solution :

Here f(x) and h(x) are equal.

L(x)  =  f(x0) + f'(x0) (x-x0)

h(x)  =  x/(x+1) and x0  =  1

h(x0)  =  1/(1+1)  =  1/2

h'(x)  =  [(x+1)(1)-x(1)] / (x+1)2

h'(x)  =  (x+1-x) / (x+1)2

h'(x)  =  1/(x+1)2

h'(x0)  =  1/(1+1)2

h'(x0)  =  1/4

By applying the values of h(x0) and h'(x0) in (1)

L(1)  =  (1/2) + (1/4)(x-1)

L(1)  =  (1/2) + (x/4)-(1/4)

L(1)  =  (1/4)(1+x) Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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