How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x) = x^2 + 4x + 2# on the interval [-3,-2]?

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Answer 1 · 2016-10-24T18:44:03

the value of #c=-2.5#

let #c∈[-3,-2]#

Then #f'(c)# can be calculated in 2 ways

The first way by differentiating #f(x)# as the function is continuous and differentiable on the interval [-3,-2]

#f'(x)=2x+4# so #f'(c)=2c+4#

#f(-3)=9-12+2=-1#
#f(-2)=4-8+2=-2#
The second is from the definition of #f'(c)#

#f'(c)=(f(-3)-f(-2))/(-3+2)=(-1+2)/(-1)=-1#

therefore, #2c+4=-1# #=>##c=-2.5#

and #c∈[-3,-2]#