Given the function #f(x)=-(6x+24)^(2/3)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [-4,-1] and find the c?

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Answer 1 · 2016-07-28T20:37:10

See below.

The Mean Value Theorem has two hypotheses:

H1 : #f# is continuous on the closed interval #[a,b]#

H2 : #f# is differentiable on the open interval #(a,b)#.

In this question, #f(x) = -(6x+24)^(2/3)# , #a=-4# and #b=-1#.

We can apply the Mean Value Theorem if both hypotheses are true.

This function is continuous on its domain, #(-oo, oo)#, so it is continuous on #[-4, -1]#

#f'(x)=-4(6x+24)^(-1/3)# which exists for all #x != 4# (for all #x in (-4,-1)#,

So #f# is differentiable on #(-4, -1)#

Therefore #f# satisfies the hypotheses of The Mean Value Theorem on the interval: #[-4, -1]#

So we know that there is a #c# in #(-4,-1)# for which

#f'(c) = (f(-1)-f(-4))/(-1-(-4))#.

To find the #c#, do the arithmetic and solve

#-4(6x+24)^(-1/3) = -(18)^(2/3)/3#

to get #x = -28/9 = -3 1/9#