How do you find the value(s) of c guaranteed by the Mean Value Theorem for Integrals for the function #f(x)= 6 cos x# on the interval# [-pi/2, pi/2]#?

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Answer 1 · 2015-08-30T15:08:34

Solve #f'(x) = (f(pi/2) - f(-pi/2))/((pi/2) - (-pi/2))# for values in # (-pi/2, pi/2)#

I would begin by checking that the hypotheses are true.

This function is continuous on # [-pi/2, pi/2]# and differentiable on # (-pi/2, pi/2)#, so the hypotheses of the Mean Value Theorem are true.

The conclusion is that there is a number #c# in # (-pi/2, pi/2)# with

#f'(c) = (f(pi/2) - f(-pi/2))/((pi/2) - (-pi/2))# .

We have been asked to find the number(s) whose existence the theorem asserts.

For #f(x) = 6cosx#, we have #f(pi/2) = f(-pi/2) = 0#, so we need to solve

#f'(c) = 0#

Furthermore #f'(x) = -6sinx# which is #0# at integer multiples of #pi#

The only integer multiple of #pi# in # (-pi/2, pi/2)# is #0 pi = 0#

#c = 0#