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

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

Step-by-step explanation

Given;

`f\mleft(x\mright)=(1)/(x)`

Now, evaluate its first derivative;

`f^(\prime)(x)=-(1)/(x^2)`

Now;

`\begin{gathered} x=0.2=(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} (1)/(0.202)=5.25(0.202-0.2) \\ =5-0.05 \\ \cong4.95 \end{gathered}`