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 $T(x)=f(a)+f^{\prime }(a)\cdot (x-a)$

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 $f^{\prime }(x)=\frac{1}{2\sqrt{x}}$, thus $f^{\prime }(64)=\frac{1}{2\cdot 8}=\frac{1}{16}$.

Therefore the linear approximation line is:

$\begin{array}{rl}T(x)=& f(a)+f^{\prime }(a)\cdot (x-a)\\ =& f(64)+f^{\prime }(64)\cdot (x-64)\\ =& 8+\frac{1}{16}\cdot (x-64)\end{array}$

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: $\frac{133}{16}$