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I have the following question:

An approximation of the value of $f(x)$ can be given by the formula $f(x)=f(x_0)+f'(x_0) \Delta x $ where $x = x_0 + \Delta x$. The formula is a consequence of the definition of the derivative.

(a) using this formula, find an approximation for $\sqrt{66}$.

(b) the approximation is likely to be improved if we replace $f'(x_0)$ by the average of $f'(x)$ and $f'(x_0)$ in the above formula. Write a formula for $f(x)$ using this average value of $f'$ rather than $f'(x_0)$.

(c) in order to transform the formula in (b) to a useful formula for finding an approximation of $f(x)$ we need to find a way to express $f'(x)$ solely in terms of $x_0$. We may approximate $f'(x)$ using the formula $f'(x) = f'(x_0) + f''(x_0) \Delta x $. Sub this approximation in for $f'(x)$ in the formula you found in b and simplify to find an approximation solely in terms of $x_0$.

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    i rlly need you to format your post. Perhaps with paragraphs and mathjax: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-01-06

1 Answers 1

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a)$$\sqrt{66}\approx\sqrt{64}+\frac1{2\sqrt{64}}(66-64)=8+\frac18=\frac{65}8$$

b)$$f(x)\approx f(x_0)+\frac{f'(x)+f'(x_0)}2(x-x_0)$$

c)$$\sqrt{66}\approx\sqrt{64}+\frac{f'(66)+\frac1{\sqrt{64}}}{4}(66-64)$$

$$f'(66)\approx\frac1{2\sqrt{64}}-\frac1{4\sqrt{64}^3}(66-64)=\frac1{16}-\frac1{512}=\frac{31}{512}$$

$$\sqrt{66}\approx8+\frac{\frac{31}{512}+\frac18}2=\frac{8287}{1024}$$

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    thank u but why is it sqrt 64 – 2017-01-06
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    @jenna That is the closest perfect square to $66$. :-) – 2017-01-06