At what x value on the interval [-2,3] does the graph of f(x)=x^2+2x-1 satisfy the mean value theorem?

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neela | High School Teacher | (Level 3) Valedictorian

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f(x) = x^2+2x-1.

To verify mean value theorem.

f(-2) = (-2)^2+2(-3)-1 = 4-6-1 = -3

f(3) = (3)^2+2(3)-1 = 9+6-1 = 14.

The mean value theorem staes that  if f(x) is a continuous function in a closed interval (a,b) and derivable in the open interval (a,b), then there exists a c in the interval (a,b) such that

f'(c) = (f(b)-f(a))/(b-a)...........(1)

a= -2 b= 3. So (f(3)-f(-2))/(3-(-2)) = (14-(-3))/(3+2) = 17/5 = 3.4.

f'(x) = (x^2+2x-1)' = 2x+3.

Therefore f'(x) = (f(3)-f(-2)/(3-(-2)) gives 2x+3 = 3.4. Or x = (3.4-3)/2 = 0.4/2 = 0.2.

Thus  if  if f(x) = x^2+2x-1 over the closed interval (-2 ,3) , then the point x= 0.2 in the interval (-2,3) satisfies  f'(x) = (f(b)-f(a))/(b-a)  in accordance with the mean value theorem.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The Mean Value Theorem states that if a function is continuous and differentiable over the closed interval [a,b], then there is a point c, that belongs to [a,b], such that:

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

To calculate f'(c), we'll have to differentiate the function f(x).

f'(x) = (x^2+2x-1)'

f'(x) = 2x + 2

Now, we'll substitute x by c:

f'(c) = 2c + 2

From the mean value theorem, we'll get:

f'(c) = [f(3) - f(-2)]/(3 + 2)

We'll calculate f(3):

f(3) = 3^2+2*3-1

f(3) =9+6-1

f(3) =14

We'll calculate f(-2):

f(-2) = (-2)^2+2*(-2)-1

f(-2) = 4 - 4 - 1

We'll eliminate like terms:

f(-2) = -1

f'(c) = (14+1)/5

f'(c) = 15/5

f'(c) = 3 (1)

But f'(c) = 2c + 2 (2).

We'll substitute (1) in (2):

2c + 2 = 3

2c = 3-2

2c = 1

c = 1/2

For x = 1/2, the mean value theorem is satisfied for the function f(x) = x^2+2x-1.

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