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

1 Answer
Jun 22, 2018

The answer is #c=1.75#

Explanation:

The mean value theorem states that if a function #f(x)# is continous on the interval #[a,b]# and differentiable on the interval #(a,b)#, then there exists a point #c# in the interval such that

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

Here,

#f(x)=-(-2x+6)^(1/2)=-sqrt(-2x+6)#

The domain is #x in (-oo, 3]#

The function is continuous and differentiable on the interval #(-2,3)#

Therefore,

#f'(x)=-1/(2sqrt(-2x+6))*-2=1/sqrt(-2x+6)#

#f'(c)=1/sqrt(-2c+6)#

#f(-2)=-sqrt(4+6)=-sqrt10#

#f(3)=-sqrt(-6+6)=0#

Therefore,

#1/sqrt(-2c+6)=(0-(-sqrt10))/(3-(-2))#

#=>#, #1/sqrt(-2c+6)=sqrt10/5#

#=>#, #sqrt(-2c+6)=5/sqrt10#

Squaring both sides

#=>#, #-2c+6=25/10=2.5#

#2c=6-2.5=3.5#

#c=3.5/2=1.75#