Linear Approximation Of Functions

Linear Approximation Of Functions
Given a function $f(x)$,if $x=a$ then the linear approximation of $f(x)$ is given by $L(x)=f'(a)(x-a)+f(a)$

Examples
Use Linear Approximation to evaluate the following functions.
(1) $f(x)=\sqrt[3]{x}$ at $x=64$ to approximate $\sqrt[3]{50}$
(2) $f(x)=\ln{x}$ at $x=1$ to approximate $f(1.5)$

Solution:

(1) $$\begin{align*} &f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}\Rightarrow{f'(x)}=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}}\\ &\text{since}\ x=a\ \text{then}\ a=64, \text{and}\ f(a)=\sqrt[3]{64}=4\\ &f'(a)=\frac{1}{3(64)^{\frac{2}{3}}}=\frac{1}{48}\\ &\text{since we have found all the necessary values for the formula,}\\ &\text{we proceed to substituting them into the formula}\\ &L(x)=f'(a)(x-a)+f(a)\\ &L(x)=\frac{1}{48}(x-64)+4\\ &L(x)=\frac{x}{48}-\frac{64}{48}+4\\ &L(x)=\frac{x}{48}+\frac{8}{3}\\ &\text{and that's it,now we need to evaluate}\ L(50)\\ &L(50)=\frac{50}{48}+\frac{8}{3}\\ &=\frac{50+128}{48}=3.7083\\ &\simeq{3.71}\\ &\text{Now we approximate}\ L(50)\ \text{to}\ \sqrt[3]{50}\\ &\simeq{3.68} \end{align*}$$

(2) $$\begin{align*} \text{using} L(x)=f'(a)(x-a)+f(a)\\ \text{since}\ x=a,\ f'(x)=\frac{1}{x}\ \text{and}\ f'(a)=\frac{1}{1}=1\ \text{because}\ a=1\\ f(a)=l\ln(1)=0\\ L(x)=(1)(x-1)+0=x-1\\ f(1.5)=L(1.5)=1.5-1=0.5 \end{align*}$$