Calculus A Question?

Question

Using the MVT(Mean Value Theorem), find the interval that conains c, so that f'(c)=0 for the function f(x)=1/(1+x^2)

Answer 1

f '(x) = 2x/(1+x^2)^2 --> f '(0) = 0
The interval is (-infinity, infinity) which contains the value c = 0.

Answer 2

f' (x) = -2x/(1+x^2)^2 is = 0 only at x = 0. So c=0 and any interval containing c=0 will work, for example [-1,1].

Answer 3

Since f(x) = /(1+x^2)

f'(x) = -2x/((1 + x^2)^2). So, f'(c) = 0 if, and only if, c = 0. It's immediate, we don't need MVT here.

Answer 4

Doing a straight out differentiation -

f(x)=1/(1+x^2)
f'(x) = - 2x /(1+x^2)^2

Setting f' = 0, solve for x=c

- 2x /(1+x^2)^2 = 0

Since (1+x^2)^2 cannot be zero for all values of x;
x must be equal to zero. Therefore c =0. Interval [0,0]

MVT:
f' =[ f(x+h) - f(x)]/h as h --> 0
f' = [1/(1+(x+h)^2) - 1/(1+x^2)]/h
f' = [(1+x^2) - (1+x^2+2xh+h^2)]/[h (1+x^2)(1+x^2+2xh+h^2)]

f' = [1+x^2-1-x^2-2xh-h^2]/h
--------------------------------------------------------------- as h --->0
(1+h^2+2hx+(2+h^2)x^2 + 2hx^3 + x^4)

(not sure if I did the expansion correctly)