The linearization is the best linear approximation Suppose

QUESTION:

The linearization is the best linear approximation Suppose that \(y = f(x)\) is differentiable at \(x = a\) and that \(g(x) = m(x ⎯ a) + c\) is a linear function in which m and c are constants. If the error \(E(x) = f(x) ⎯ g(x)\) were small enough near \(x = a\), we might think of using g as a linear approximation of f instead of the linearization \(L(x)=f(a)+f^{\prime}(a)(x-a)\). Show that if we impose on g the conditions

1. \(E(a) = 0\)        The approximation error is zero at \(x = a\).

2.\(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\)        The error is negligible when compared with \(x-a\).

then \(g(x)=f(a)+f^{\prime}(a)(x-a)\). Thus the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x = a\)and negligible in comparison with \(x-a\).

Equation Transcription:

Text Transcription:

y = f(x)

x=a

g(x) = m(x-a) + c

E(x) = f(x)-g(x)

L(x) = f(a) + f’(a)(x-a)

E(a) = 0

lim_x ->aE(x/x-a=0