Another Linear approximation question

In summary, the conversation discusses using the Linear Approximation method to estimate Δf and calculating the error and percentage error using a calculator. The derivative of the given function is first obtained, and then the linear approximation is found by plugging in the given values. The error is then calculated by finding the difference between the estimated value and the actual value, and the percentage error is found by dividing the error by the actual value. The conversation ends with a clarification on the error calculation.
  • #1
bcahmel
25
0

Homework Statement


stimate Δf using the Linear Approximation and use a calculator to compute both the error and the percentage error.
f(x) =1/(1+x^2) , a = 3, Δx = 0.5

Homework Equations



f'(a)(x)
percentage error= abs(error) divided by actual value

The Attempt at a Solution


So first I got the derivative which is -2x/(1+x)^2.
Then I plugged the a value, 3 into it which came out to be -3/50 and then multiplied it by x, 0.5, to get my linear approximation of -0.3. I understand this part.
Now to find the error, I first have to find the actual value on the calculator. So I plugged 3.5 into the original function, 1/(1+3.5^2)= 1/13.25. Is this right so far?
Plugging just 3 into the equation I get 1/10.

Now 1/13.25- 1/10 is about -.0245. This is the error, I think...

and percentage error would be .0055/-.0245=22.44% Seems high..

Basically I'm confused on the error part. I would be really grateful if anyone could straighten out how to do this!
 
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  • #2
shouldn't the derivative be -2x/(1+x^2)^2
 
  • #3
The linear approximation gives you an estimate of f(3.5), so the error you want to calculate is

[tex]\frac{f_\mathrm{est}(3.5)-f(3.5)}{f(3.5)}[/tex]
 
  • #4
thanks lanedance, I typed it wrong on the computer- but you're right that is the derivative.

And vela, that's makes sense. So I think I did the error part right!
 
  • #5
The error was something like 7% or so, not 22%, so you should recheck what you did.
 

FAQ: Another Linear approximation question

1. What is linear approximation?

Linear approximation is a method used in calculus to estimate the value of a function at a certain point by using the slope of the tangent line at a nearby point. It is often used to simplify complicated functions and make them easier to work with.

2. How is linear approximation calculated?

The linear approximation formula is y = f(a) + f'(a)(x-a), where a is the point at which the function is being approximated and f'(a) is the derivative of the function at that point. This formula gives the equation of the tangent line at point a, which can be used to estimate the value of the function at a nearby point.

3. In what situations is linear approximation useful?

Linear approximation is useful in situations where a function is too complicated to evaluate directly, but can be approximated by a simpler linear function. It is commonly used in physics, engineering, and economics to simplify mathematical models and make them easier to analyze.

4. What are the limitations of linear approximation?

Linear approximation is only accurate for small intervals around the point at which the function is being approximated. The further away the point is from the original point, the less accurate the approximation will be. Additionally, linear approximation cannot be used for functions with sharp curves or discontinuities.

5. How is linear approximation different from linear regression?

Linear approximation is a mathematical technique used to approximate the value of a function at a specific point, while linear regression is a statistical method used to analyze the relationship between two variables. Linear regression involves finding the best-fit line for a set of data points, while linear approximation involves finding the equation of a tangent line for a single point on a function.

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