How do you find the point c in the interval #-1<=x<=1# such that f(c) is equation to the average value of #f(x)=x^2-x#?

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Answer 1 · 2017-08-01T14:26:16

Find the average value. Set the function equal to that value. Solve the equation in the interval. #c = 1/2 - sqrt21/6#

Average value on #[-1,1]#:

#bar(y) = 1/(1-(-1)) int_-1^1 (x^2-x) dx#

# = 1/2 [x^3/3-x^2/2]_-1^1#

# = 1/3#

Solve

#x^2-x = 1/3#

#3x^2-3x-1=0#

#x = (3+-sqrt(9+12))/6#

# = 1/2 +- sqrt21/6#

#5 < sqrt21 < #6, so #1/2 < sqrt21/6 < 1#. Thus

#1/2 + sqrt21/6# is outside the interval and

#1/2 - sqrt21/6# is in the interval #[-1,1]#.

#c = 1/2 - sqrt21/6#