Answer
The linearization of $f(x)$ at $x=a$ is also the quadratic approximation of $f(x)$ at $x=a$ .
Work Step by Step
We know that the linearization of $f(x)$ at $x=a$ can be defined as:
$L(x) =f(x)=f(a)+(x-a) +f’(a)$
And the quadratic approximation of $f(x)$ at $x=a$ is as follows:
$Q(x) =f(x)$
or, $=f(a)+(x-a) +f’(a)+\dfrac{(x-a)^2}{2!}f’’(a)$
Since, $f’’(a)=0$
$\implies Q(x) = f(a)+(x-a) +f’(a)+\dfrac{(x-a)^2}{2!}f’’(a) \implies L(x)=Q(x)$
So, the linearization of $f(x)$ at $x=a$ is also the quadratic approximation of $f(x)$ at $x=a$ .
