Question

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

Answer 1

The linearization
`\( l(x) \)`of the function
`\( f(x) = x^4 - 2x^2 \)`at
`\( a = 1 \)` is given by
`\( l(x) = -x + 3 \).`

Step-by-step explanation:

To find the linearization of a function at a specific point, we use the formula
`\( l(x) = f(a) + f'(a)(x - a) \)`, where
`\( f'(x) \)` is the derivative of
`\( f(x) \).`First, we find the derivative of
`\( f(x) \)` with respect to
`\( x \)`, which is
`\( f'(x) = 4x^3 - 4x \).`

Next, we evaluate
`\( f \)`and
`\( f' \)`at the given point
`\( a = 1 \). \( f(1) = 1^4 - 2 \cdot 1^2 = -1 \),` and
`\( f'(1) = 4 \cdot 1^3 - 4 \cdot 1 = 0 \).`

Now, we substitute these values into the linearization formula:

`\[ l(x) = -1 + 0 \cdot (x - 1) \]`

Simplifying, we get
`\( l(x) = -x + 3 \).`

Therefore, the linearization
`\( l(x) \)`of the function
`\( f(x) = x^4 - 2x^2 \)` at
`\( a = 1 \)` is
`\( l(x) = -x + 3 \).` This linearization is an approximation of the original function near the point
`\( x = 1 \)`and is useful in estimating function values close to that point.

Answer 2

The linearization l(x) of the function f(x) = x⁴ - 2x² at a = 1 is y = -1. This is obtained by finding the derivative, evaluating it at x = 1, and determining the value of the function at that point.

Step-by-step explanation:

To find the linearization l(x) of the function f(x) = x⁴ - 2x² at a = 1, we need to determine the tangent line to the function at that point.

The equation for a tangent line in linearization is given by l(x) = f(a) + f'(a)(x - a).

First, we find the derivative of the function, f'(x), which is 4x³ - 4x. Now, we evaluate this derivative at a = 1: f'(1) = 4(1)³ - 4(1) = 0.

Then we find the value of the function at that point: f(1) = 1⁴ - 2(1)² = -1.

With these values, the linearization equation becomes l(x) = -1 + 0·(x - 1) = -1.

Therefore, the linearization of the function at a = 1 is simply y = -1.