Linearization of a Function at a Point

Problem 168: Find the linearization of $f(x)$ at $x=a$ :

(1) $\begin{equation*} f(x) = x + \frac{1}{x}, \qquad a = 1\end{equation*}$

Solution: The linearization of a differentiable function at a point is given by

(2) $\begin{equation*}L(x) = f(a) + f'(a)(x-a). \end{equation*}$

Hence,

(3) $\begin{align*}f(a) & = f(1)= 1 + \frac{1}{1} = 1 + 1 = 2, \ \text{and} \\ f'(a) & = f'(1) = 1-\frac{1}{1^2}=1 - 1 = 0. \end{align*}$

Therefore,

(4) $\begin{align*} L(x) & = f(a) + f'(a) (x - a) \\ & = 2 + 0(x-1) \\ & = 0x - 0 + 2 \\ & = 2.\end{align*}$

That is,

(5) $\begin{equation*} L(x) = 2 \end{equation*}$

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