Question

square root of 20.591

Answer 1

use a calculator >.< type in 20.951 then the square root number

Answer 2

Lets first do a linear approximation of it.

This is \(\large\color{black}{ \displaystyle f(x)=\sqrt{x} }\), and value we are going to use is \(a=25\).

In other words, we are drawing a tangent line (at x=25) to this function, and use this line to estimate the value of \(\sqrt{20.951}\). (This process is also known as \(\normalsize \color{blue}{\rm linearization.}\))

\(\large\color{black}{ \displaystyle L(x)=f(a)+f'(a)(x-a) }\)
(this is just the tangent line, to a curve f(x), at x=a, in general)

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\(\large\color{black}{ \displaystyle \frac{d}{dx}\sqrt{x} ~=\frac{d}{dx}x^{\LARGE ½}~=\frac{1}{2}x^{{\LARGE ½}-1}~=\frac{1}{2}x^{-\LARGE ½} ~=\frac{1}{2\sqrt{x}} }\)


\(\large\color{black}{ \displaystyle f(a)~=\sqrt{a}~=\sqrt{25}=5 }\)

\(\large\color{black}{ \displaystyle f'(a)~=\frac{1}{2\sqrt{a}} ~=\frac{1}{2\sqrt{25}}~=\frac{1}{10} }\)

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\(\large\color{black}{ \displaystyle L(x)=5+\frac{1}{10}(x-25) }\)

Now, you can use this equation to approximate your square root (for values near x=25)

\(\large\color{black}{ \displaystyle L(20.951)=5+\frac{1}{10}(\color{red}{20.951} -25) }\)

\(\large\color{black}{ \displaystyle L(20.951)=5+\frac{1}{10}(-4.049) }\)

\(\large\color{black}{ \displaystyle L(20.951)=5-0.4049 }\)

\(\large\color{black}{ \displaystyle L(20.951)=4.5951 }\)

Answer 3

You can get a beter approximation, by drawing a bigger polynomial ....

\(\color{black}{ \displaystyle f(a)+f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3 ....}\)

Answer 4

Or, this can be written entirely as:

\(\color{black}{ \large \displaystyle \sum_{{\rm n=0}}^{\infty}\frac{\left(\displaystyle \frac{d^{\rm n}~f(x)}{dx^{\rm n}}~{\Huge |}_{x=a}\right) }{{\rm n}!}\times (x-a)^{\rm n}}\)

More traditionally written as:

\(\color{black}{ \large \displaystyle \sum_{{\rm n=0}}^{\infty}\frac{f^{\rm n}(a) }{{\rm n}!}\times (x-a)^{\rm n}}\)