Question

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

Answer 1

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]}