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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

Use linear approximation ie the tangent line to approximate as follows Let fx and find the equation of the tangent line to fx at a nice point near 0202 Then use class=

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ANSWER

[tex]\begin{equation*} 4.95 \end{equation*}[/tex]

EXPLANATION

Given;

[tex]f\mleft(x\mright)=\frac{1}{x}[/tex]

Now, evaluate its first derivative;

[tex]f^{\prime}(x)=-\frac{1}{x^2}[/tex]

Now;

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

Hence, we have;

[tex]l(x)=5-25(x-0.2)=5-25x+5[/tex]

by substitution;

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

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