Linear Approximation of \sqrt[3]{27.02} using f(x)+f'(x)(x-a) method

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In summary, the conversation discusses finding the linear approximation of \sqrt[3]{27.02} using the equation f(x)+f'(x)(x-a). The approach used was to work with \sqrt[3]{27} as an easily known value, with f(x)=x^{1/3} and f'(x)=1/3x^{-2/3}. The resulting approximation was calculated to be 3.00074, and a sanity check was recommended to verify the accuracy of the approximation.
  • #1
forestmine
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Homework Statement



Find the linear approximation of [itex]\sqrt[3]{27.02}[/itex]

Homework Equations



f(x)+f'(x)(x-a)


The Attempt at a Solution



So what I did was work with[itex]\sqrt[3]{27}[/itex] since that's an easily known value. So my f(x)=x[itex]^{1/3}[/itex] and my f'(x)=1/3x[itex]^{-2/3}[/itex]. From there, I worked f(27) = 3, and f'(27)=1/27.

Then I used the above equation. 3+1/27(x-27). For my value of x, I used 27.02, and got .0004.

Something tells me I'm doing something incorrectly, though...

Thanks for the help!
 
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  • #2
Are you saying 3+1/27(x-27) = .0004? It has to larger than 3.
 
  • #3
My mistake -- calculated that incorrectly. I actually get 3.00074. Was my method correct in that case?
 
  • #4
forestmine said:
My mistake -- calculated that incorrectly. I actually get 3.00074. Was my method correct in that case?

Your method looks ok to me. A good sanity check is to calculate the cube root of 27.02 and compare to the approximation.
 
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  • #5
Thank goodness for sanity checks. Thanks a lot!
 

FAQ: Linear Approximation of \sqrt[3]{27.02} using f(x)+f'(x)(x-a) method

What is a linear approximation?

A linear approximation is an approximation of a nonlinear function by a linear function. It involves finding the equation of a line that best fits the curve of a nonlinear function at a specific point, in order to estimate the value of the function at that point.

How is a linear approximation calculated?

A linear approximation is calculated using the tangent line to the curve of a nonlinear function at a specific point. The slope of the tangent line at that point becomes the slope of the linear approximation, and the y-intercept of the tangent line is used to find the y-value of the linear approximation at that point.

What is the purpose of using linear approximations?

The main purpose of using linear approximations is to simplify complex nonlinear functions and make them easier to work with. They can also provide a close estimate of the actual value of the function at a specific point, which can be useful in various mathematical and scientific applications.

Can linear approximations be used for any type of function?

No, linear approximations can only be used for differentiable functions, which means that the function must have a defined derivative at the point of approximation. Non-differentiable functions, such as those with vertical tangent lines, cannot be approximated linearly.

How can linear approximations be improved?

Linear approximations can be improved by using higher-order approximations, such as quadratic or cubic approximations, which involve fitting curves instead of straight lines. Additionally, using smaller intervals and more data points can also lead to more accurate linear approximations.

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