MHB Dman's question at Yahoo Answers concerning linear approximates

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The discussion focuses on using linear approximation to estimate 1/0.101 by finding the tangent line of the function f(x) = 1/x at a nearby point. The derivative of f(x) is calculated as -1/x², evaluated at x = 0.1, to derive the equation of the tangent line. By applying the linear approximation formula, the estimate for 1/0.101 is found to be approximately 9.9. This method provides a close approximation compared to the actual value of 9.900990099009... The discussion emphasizes the effectiveness of linear approximation in estimating values of functions near known points.
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Here is the question:

Use linear approximation, Let 1/.101 and f(x)=1/x and find the equation of the tangent line?

Use linear approximation, i.e. the tangent line, to approximate 1/.101 as follows: Let f(x)=1/x and find the equation of the tangent line to f(x) at a "nice" point near .101 Then use this to approximate 1/.101

Here is a link to the question:

Use linear approximation, Let 1/.101 and f(x)=1/x and find the equation of the tangent line? - Yahoo! Answers

I have posted a link there to this topic so the OP may find my response.
 
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Hello dman,

I would begin with:

$\displaystyle \frac{\Delta f}{\Delta x}\approx\frac{df}{dx}$

Using $\Delta f=f(x+\Delta x)-f(x)$ and multiplying through by $\Delta x$ we obtain:

$\displaystyle f(x+\Delta x)\approx\frac{df}{dx}\Delta x+f(x)$

Now, using the following:

$\displaystyle f(x)=\frac{1}{x}\,\therefore\,\frac{df}{dx}=-\frac{1}{x^2},\,x=0.1,\,\Delta x=0.001$

we may state:

$\displaystyle \frac{1}{0.101}\approx-\frac{1}{0.01}\cdot0.001+\frac{1}{0.1}=10-0.1=9.9$

For comparison:

$\displaystyle \frac{1}{0.101}=9.900990099009...$.
 
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