How do you determine all values of c that satisfy the conclusion of the mean value theorem on the interval [0, 8] for #f(x) = x^3 + x - 1#?
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Solve the equation #f'(x) = (f(8)-f(0))/(8-0)#. Every solution in #(0,8)# is a permissible value of #c#.
#f'(x) = 3x^2+1#.
#(f(8)-f(0))/(8-0) = 65#.
#3x^2+1 = 65# if and only if #x= +- 8/sqrt3#. The negative solution is not in the interval.
The only value of #c# that satisfies the conclusion is #8/sqrt3#.
The mean value theorem states: #(f(b)-f(a))/(b-a)#
so you have #(f(8)-f(0))/(8-0)#
Plug 8 and 0 into the equation #f(x)= x^3 + x -1 #
f(8)= 512 + 8 - 1 = 519
f(0)= 0 + 0 - 1 = -1
#(519+1)/(8-0)# = 65
Now take the derivative of the function #f(x)= x^3 + x -1 #
#3x^2+1# is the derivative, now set it equal to 65.
#3x^2+1#= 65, subtract the 1 to the other side
#3x^2#= 64, divide the 3 to the other side
#x^2#= #64/3#, take square of both side
x= #sqrt(64)/sqrt(3)#
x= #+-8/sqrt(3)#
