Question
question:
In the same manner as in the Week 6 homework 6.1. #2. we can show that g ii differentiable at 0,g'(0)= 0 Find the derivative g' (x) for x nonzero, x in [-1, 1). You may apply the usual calculus differentiation rules such as the product rule, the chain rule, and the derivatives of trig functions. (Note that the rules were among the theorems in our differentiation chapter.) Simplify as much as possible. (If you are rusty and need practice, there is some practice using the chain rule in at the end of section 6.1 or consult your calculus book.) looking at the (a) result, carefully explain how you know that g' is not bounded on the interval [-1,1). Can the Fundamental Theorem of Calculus II be applied on the interval [-1,1). with g as above and f= g'?, (The theorem is stated above # lpart (h), Why or why not? Briefly explain.
Explanation / Answer
1)
a)3*x^2*sin(1/x^3) - (3*cos(1/x^3))/x
b)as x goes to 0
second part of above expression goes to negitve infinity and hence it is unbounded .
c) no we cannot appy the funadamental theorem as the function goes unbounded (f=g' is not finite in the intreval) and hence cannot appky the theorem. as f is NOT continuous at '0'
