Science:Math Exam Resources/Courses/MATH104/December 2016/Question 04 (a)/Solution 1

Note that The linear approximation of $f(x)$ at a point $x=a$ is given by $T(x)=f(a)+f'(a)\cdot (x-a)$

To estimate ${\sqrt {69}}$, we first find the linear approximation of $f(x)={\sqrt {x}}$ at $a=64$. (The reason we choose $64$ is explained in the Hint)

For this purpose, we need $f'(x)={\frac {1}{2{\sqrt {x}}}}$, thus $f'(64)={\frac {1}{2\cdot 8}}={\frac {1}{16}}$.

Therefore the linear approximation line is:

${\begin{aligned}T(x)=&f(a)+f'(a)\cdot (x-a)\\=&f(64)+f'(64)\cdot (x-64)\\=&8+{\frac {1}{16}}\cdot (x-64)\end{aligned}}$

Since $f(x)\sim T(x)$ around $x$ near $64$, we have

${\sqrt {69}}=f(69)\sim T(69)=8+{\frac {1}{16}}\cdot (69-64)=8+{\frac {5}{16}}={\frac {133}{16}}$.

Answer: $\color {blue}{\frac {133}{16}}$