Raymond B. answered 02/05/23
Math, microeconomics or criminal justice
f(x)= x^3/4 = (x^3)/4
4y = x^3
on the interval [0,1]
y=0 for x=0
y = 1/4 for x=1
f(c) = c^3/4
f'(c) = 3c^2/4
f'(c) = (f(1)-f(0)]/(1-0) = 1/4= 3c^2/4
3c^2/4 = 1/4
3c^2 =1
c^2 = 1/3
c = sqr(1/3) = 1/sqr3
c= sqr3/3
= about 1.732.../3
= about .577350269189626
c= sqr3/3 = about .577