Question #f6ebb

Nov 16, 2014

The linearization function is $L \left(x\right) = - 9 - 16 x$

For any function f(x), the linearization function L(x) at any point a can be found using the relation

$L \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right)$
where f'(x) is the first derivative of the function f(x).

Basically, this formula for $L \left(x\right)$ is the value of the function at the point a plus the slope of the function at the point a times any displacement from the point a.

Using information from the question:
$f \left(x\right) = {x}^{4} + 6 {x}^{2}$ and $a = - 1$
$f \left(a\right) = f \left(- 1\right) = {\left(- 1\right)}^{4} + 6 {\left(- 1\right)}^{2} = 1 + 6 = 7$
$f ' \left(x\right) = 4 {x}^{3} + 12 x$
$f ' \left(a\right) = f ' \left(- 1\right) = 4 {\left(- 1\right)}^{3} + 12 \left(- 1\right) = - 4 - 12 = - 16$
$L \left(x\right) = f \left(a\right) + f ' \left(a\right) \left(x - a\right) = 7 - 16 \left(x - \left(- 1\right)\right)$
which simplifies to
$L \left(x\right) = - 9 - 16 x$

As a check, we can verify that $L \left(a\right) = - 9 + 16 = 7 = f \left(a\right)$