Find the average value f ave of f ( x ) = x ^5 between -1 and 1, then find a number c in [-1,1] where f ( c ) = f ave .

Question

Find the average value  of  between -1 and 1, then find a number  in [-1,1] where .

Umm I am very confused...where do i start?

Answer 1

We have to find integral average: v_av = 1/(1 - (-1))∫_-1¹x⁵dx = 1/2·x⁶/6_-1¹ = 0;

f(c) = c⁵ = 0, c = 0.

Answer 2

The average of the function on [a,b] is [ f(b)-f(a) ]/ (b-a)

a=-1 b = 1 because the interval is [-1,1]

f(1) = 1^5 = 1

f(-1) = (-1)^5 = -1

plugging into the formula: [ 1 - -1]/(1 - -1) = (1 + +1)/(1 + +1)

= 2/2 = 1

Now you want to know WHERE the function is equal to 1 on the interval

as guaranteed by the mean value theorem

1 = C^5 = f(C) for C in [-1,1]

The FIFTH root of both sides:

1=c