I don't think you need to derive the result.Hemolymph said:Homework Statement
Use a graphing calculator or computer to verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Round the answers to two decimal places.)
tan(x) ≈ x
Homework Equations
derivative of tan(x) = sec^2(x)
The Attempt at a Solution
f(0)= tan(0)=0
fprime(0)= sec^2(0)= 1
L(x)=f(a)+fprime(a)(x-a)
L(x)=0+1(x-0)
L(x)=x
not sure how to continue from here
Linear approximation given accuracy points is a mathematical technique used to approximate the value of a function at a specific point by using the value and slope of the function at nearby known points.
This technique is useful because it allows for quick and easy estimation of a function's value at a specific point, without having to evaluate the function at every single point.
The linear approximation formula is y = f(a) + f'(a)(x-a), where a is the known point, f(a) is the known value of the function at point a, and f'(a) is the slope of the function at point a.
The accuracy of linear approximation given accuracy points depends on the distance between the known point and the point at which the function's value is being approximated. The closer the points are, the more accurate the approximation will be.
Linear approximation given accuracy points is commonly used in physics, engineering, and economics to estimate values of functions and to make predictions based on known data points.