How do you determine all values of c that satisfy the mean value theorem on the interval [1, 4] for #f(x)=1/sqrt(x)#?

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Answer 1 · 2017-08-06T12:57:35

The value of #c=2.08#

The mean value theorem states that if a function #f(x)# is continuous on the interval #[a,b]# and differentiable on the interval #(a,b)#, then

#EE c in [a,b]# such that

#f'(c)=(f(b)-f(a))/(b-a)#

Here,

#f(x)=1/sqrtx#

The domain of #f(x)# is #D_f(x) in (0, +oo)#

The interval #[1,4] in D_f(x)#

#f'(x)=-x^(-3/2)/2#

#f'(c)=-c^(-3/2)/2#

#f(4)=1/sqrt4=1/2#

#f(1)=1/sqrt1=1#

Therefore,

#-c^(-3/2)/2=(1/2-1)/(4-1)=-1/6#

#c^(-3/2)=1/3#

#c^(3/2)=3#

#c^3=9#

#c=root(3)9=2.08#

#2.08 in [1,4]#