How do you find the linearization of f(x) = x^4 + 5x^2 at the point a=1?

1 Answer
Jim H
Nov 22, 2016

The linearization is the tangent line. (Or maybe it is more helpful to say: it is a way of thinking about and using the tangent line.)

Explanation:

f(x) = x^4+5x^2

At x=1, we have y = f(1) = 6

f'(x) = 4x^3+10x so at x=1, the slope of the tangent line is m=f'(1) = 14

Equation of tangent line in point-slope form:

y-6 = 14(x-1)

Linearization at a=1 (in a form I am used to):

L(x) = 6+14(x-1)

Free example of using the linearization

f(1.1) ~~ L(1.1) = 6+14(1.1-1) = 6+14(0.1) = 6+1.4 = 7.4

(The point on the tangent line with x-coordinate 1.1 has y-coordinate 7.4.)

The exact value is f(1.1) = 1.7541

If you're very patient and steady of hand, you can find these two values using the graph:
graph{(y-x^4-5x^2)(y-14x+8)=0 [0.1604, 1.846, 6.859, 7.7024]}

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