Question

Use linear approximation, i.e. the tangent line, to approximate as follows: Let f(x) = and find the equation of the tangent line to f(x) at a nice" point near 0.202. Then use this to 1/0.202 approximate 1/0.202

Answer 1

ANSWER

$\begin{equation*} 4.95 \end{equation*}$

EXPLANATION

Given;

$f\mleft(x\mright)=\frac{1}{x}$

Now, evaluate its first derivative;

$f^{\prime}(x)=-\frac{1}{x^2}$

Now;

$\begin{gathered} x=0.2=\frac{1}{5} \\ f(x)=5,f^{\prime}(x)=-25 \end{gathered}$

Hence, we have;

$l(x)=5-25(x-0.2)=5-25x+5$

by substitution;

$\begin{gathered} \frac{1}{0.202}=5.25(0.202-0.2) \\ =5-0.05 \\ \cong4.95 \end{gathered}$