FINDING LINEAR APPROXIMATION FOR A FUNCTION

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

x₀ ∈ (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)

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)

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