How do you find the point c in the interval #0<=x<=3# such that f(c) is equation to the average value of #f(x)=3x^2#?

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Answer 1 · 2017-08-06T22:18:55

Find the average value. Set #f(x)# equal to the average value. Solve that equation in the interval #[0,3]#.

#f_"ave" = 1/(3-0) int_0^3 3x^2 dx#

# = 1/3 [x^3]_0^3#

# 1/3[(3^3)-(0^3)] = 1/3(27) = 9#

Now solve

#3x^2 = 9# #" "# in #[0,3]#

#x^2=3#

#x=+-sqrt3# #" "# Note that the algebra gives us two solutions, but one of them is not in the interval.

#c = sqrt3#