Chapter 5.5, Problem 70E

To determine

To prove: that $g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)$

Explanation of Solution

Given information:

The given conditions are.

  $\begin{array}{l}\text{i}\text{. }E\left(a\right)=0\\ \text{ii}\text{. }\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=0\end{array}$

Proof:

Since, by the definition

  $E\left(x\right)=f\left(x\right)-c+\left(x-a\right)\left(f^{\prime }\left(a\right)-m\right)$

If $E\left(a\right)=0$ then

  $\begin{array}{l}f\left(a\right)-c=0\\ f\left(a\right)=c\end{array}$

Since

  $\underset{x\to a}{\lim }\frac{E\left(a\right)}{x-a}=\underset{x\to a}{\lim }\frac{f\left(a\right)-c+\left(x-a\right)\left(f^{\prime }\left(a\right)-m\right)}{x-a}$

If $\underset{x\to a}{\lim }\frac{E\left(x\right)}{x-a}=0$

  $\begin{array}{l}{f}^{\prime }\left(a\right)-m=0\\ f^{\prime }\left(a\right)=m\end{array}$

You have

  $g\left(x\right)=m\left(x-a\right)+c$

Now substitute the expressions obtained for c and m .

  $\begin{array}{l}g\left(x\right)=m\left(x-a\right)+c\\ g\left(x\right)=f^{\prime }\left(a\right)\left(x-a\right)+f\left(a\right)\left[f\left(a\right)=c,f^{\prime }\left(a\right)=m\right]\\ g\left(x\right)=f\left(a\right)+f^{\prime }\left(a\right)\left(x-a\right)\end{array}$

Thus, the linearization gives the only linear approximation whose error is both zero at $x=a$ and negligible in comparison with $\left(x-a\right)$