Linear approximation given accuracy points

Hemolymph
Messages
30
Reaction score
0

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
 
Physics news on Phys.org
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
I don't think you need to derive the result.

Graph y=tan(x) and y=x for -1 ≤ x ≤ 1 or some similar interval near zero.

Make a table of x and tan(x) for some values starting at x=0, and continuing until you have a difference of 0.1.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top