For f(x)=x^2+3x−1, find all possible values c in the interval [-2,2] that satisfy the Mean Value Theorem.

Answer 1

The Mean Value Theorem states that if f(x) in the interval [a,b] and differentiable in the interval (a,b), there exist c where a < c < b such that:

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

Given:

f(x) = x²+3x-1

a=-2

b=2

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Solve the derivative of f(x)

f'(x) = 2x+3

f'(c) = 2c+3

f(a) = f(-2) = (-2)²+3(-2)-1 = -3

f(b) = f(2) = (2)2+3(2)-1 = 9

Let's substitute all the values from f'(c) = (f(b)-f(a))/(b-a):

2c+3 = (9-(-3))/(2-(-2))

2c+3 = 12/4

2c+3 = 3

2c = 0

c = 0

There is only one value of c and that is c=0.