shamieh said:Use a linear approximation to find a good approximation to \(\displaystyle \sqrt{100.4}\)
\(\displaystyle x = 100.4\)
\(\displaystyle x1 = 100\)
\(\displaystyle y1 = 10\)
\(\displaystyle y - 10 = \frac{1}{20}(100.4 - 100) \)
\(\displaystyle y = 10.20 \)
I like Serena said:Looks correct... except for a small calculation mistake.
Did you check what $10.20^2$ is?
That should immediately reveal the mistake.
shamieh said:its 104.04 but I don't understand where I went wrong. Why can't I say 1/20 = .05 and then say .05 * .4 = .20, then finally 10 + .20 = 10.20 ?
shamieh said:10.02 is the correct answer then correct?
Linear approximation is a mathematical technique used to estimate the value of a function at a particular point by using the value of the function at a nearby point. It involves finding the equation of a line tangent to the function at the given point.
Linear approximation is important because it allows us to approximate complex functions with simpler linear equations, making it easier to analyze and calculate values. It is also used in real-world applications such as engineering, physics, and economics.
The accuracy of linear approximation depends on how close the chosen point is to the point of interest. The closer the chosen point is, the more accurate the approximation will be. However, it is important to note that linear approximation is not exact and can introduce errors, especially for functions with high curvature or at points far away from the chosen point.
Linear approximation is only accurate for small intervals and cannot be used for functions with high curvature or sharp turns. It also assumes that the function is continuous and differentiable at the chosen point. Additionally, linear approximation may not be suitable for functions with multiple variables.
Linear interpolation is a method of estimating values between known data points, while linear approximation is used to estimate the value of a function at a specific point. Linear interpolation involves finding the equation of a line between two known points and using it to find the value at a desired point, while linear approximation involves finding the tangent line at a chosen point and using it to estimate the function's value at that point.