Find the linearization l(x) of the function at a. f(x) = x⁴+ 2x², x = 1

Solution:

Sometimes we can approximate complicated functions with simpler ones that give the accuracy we want for specific applications and are easier to work with. This approximation is called Linearization and is based on tangent lines. It is based on the principle that when we draw a tangent to a curve at a point to the curve the tangent line gives good approximations to the y-values on the curve. Locally every differential curve behaves like a straight line.

So if f is differentiable at x = a then the approximating function

L(x) = f(x) + f’(a)(x - a)

is the linearization of f at a.

The approximation f(x) ≈ L(x) of f by L is the standard linear approximation of f at a.

The point x = a is the center of approximation.

f(x) = x⁴+ 2x² --- (1)

f(a) = (a)⁴ + 2a²

since a = 1 in the problem statement

f(1) = (1)⁴ + 2(1)²

f(1) = 1 + 2 = 3

Differentiating equation (1) we get

f’(x) = 4x³ + 4x

f’(1) = 4(1)³ + 4(1) = 4 + 4 = 8

Therefore the approximating function is

L(1) = f(1) + f’(1) × (x -1)

= 3 + 8(x - 1) = 3 + 8x - 8

= 3 + 8x - 8

= 8x + 5

Therefore the approximation f(x) ≈ L(x) of f by L is the standard linear approximation of f at x = 1.

The point x = 1 is the center of the approximation.

f(1) = L(1) = 8x + 5