Wednesday, January 15, 2020

Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 42

The mean value theorem is applicable to the given function, since it is a polynomial function. All polynomial functions are continuous and differentiable on R, hence, the given function is continuous and differentiable on interval.
The mean value theorem states:
f(b) - f(a) = f'(c)(b-a)
Replacing 1 for b and 0 for a, yields:
f(2) - f(-1) = f'(c)(2+1)
Evaluating f(2) and f(-1) yields:
f(2) =(2+1)/2 => f(2) =3/2
f(-1) = (-1+1)/(-1) => f(-1) = 0
You need to evaluate f'(c), using quotient rule:
f'(c) = ((c+1)/c)' => f'(c) = ((c+1)'*c - (c+1)*c')/(c^2) => f'(c) = (c - c - 1)/(c^2) => f'(c) = -1/(c^2)
Replacing the found values in equation f(2) - f(-1) = f'(c)(2+1):
3/2-0= -3/(c^2) =>1/2 =-1/(c^2)=>c^2 = -2 !in R
Hence, in this case, there is no valid c value for the mean value theorem to be applied.

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