Find c in the interval [-2,1] which satisfies the mean value theorem for f(x)=x^3-2x^2+x+1
but c=6 is not in the interval [-2,1]
you want c such that f'(c) = 6
That is,
3c^2 - 4c + 1 = 6
c = (2±√19)/3
so we want c = (2-√19)/3 ≈ -0.7863
which is in the interval [-2,1]
Apologies for the mistake. You are correct. The correct value of c that satisfies f'(c) = 6 in the interval [-2, 1] is c = (2-√19)/3 ≈ -0.7863. Thank you for catching that mistake.
To find the value of c that satisfies the mean value theorem on the interval [-2, 1], we need to follow these steps:
Step 1: Check the conditions of the mean value theorem:
The function f(x) = x^3 - 2x^2 + x + 1 is continuous on the closed interval [-2, 1] and differentiable on the open interval (-2, 1).
Step 2: Find the average rate of change of f(x) over the interval [-2, 1]:
The average rate of change of f(x) is given by the formula:
avg rate of change = (f(b) - f(a))/(b - a)
where a and b are the endpoints of the interval.
Substituting the values, we get:
avg rate of change = (f(1) - f(-2))/(1 - (-2))
Step 3: Calculate the average rate of change:
f(1) = (1)^3 - 2(1)^2 + (1) + 1 = 1 - 2 + 1 + 1 = 1
f(-2) = (-2)^3 - 2(-2)^2 + (-2) + 1 = -8 - 8 - 2 + 1 = -17
avg rate of change = (1 - (-17))/(1 - (-2)) = (1 + 17)/(1 + 2) = 18/3 = 6
Step 4: Apply the mean value theorem:
According to the mean value theorem, there exists at least one value c in the interval (-2, 1) such that f'(c) = avg rate of change.
Let's find the derivative of f(x) and solve for c:
f(x) = x^3 - 2x^2 + x + 1
f'(x) = 3x^2 - 4x + 1
Setting f'(c) = 6, we have:
3c^2 - 4c + 1 = 6
3c^2 - 4c - 5 = 0
Step 5: Solve for c:
We can solve the quadratic equation using factoring, completing the square, or quadratic formula. Let's use the quadratic formula:
c = (-b ± √(b^2 - 4ac))/(2a)
For our equation 3c^2 - 4c - 5 = 0, the values of a, b, and c are:
a = 3
b = -4
c = -5
Substituting these values into the formula, we get:
c = (-(-4) ± √((-4)^2 - 4(3)(-5)))/(2(3))
c = (4 ± √(16 + 60))/(6)
c = (4 ± √(76))/(6)
c = (4 ± √(19))/3
So, the values of c that satisfy the mean value theorem for the function f(x) = x^3 - 2x^2 + x + 1 on the interval [-2, 1] are (4 + √19)/3 and (4 - √19)/3.
To apply the mean value theorem, we need to find a value c in the interval [-2, 1] such that f'(c) = (f(1) - f(-2))/(1 - (-2)).
First, let's find f'(x):
f'(x) = d/dx (x^3 - 2x^2 + x + 1)
= 3x^2 - 4x + 1
Next, let's find f(1) and f(-2):
f(1) = (1)^3 - 2(1)^2 + (1) + 1
= 1 - 2 + 1 + 1
= 1
f(-2) = (-2)^3 - 2(-2)^2 + (-2) + 1
= -8 - 8 - 2 + 1
= -17
Now, let's evaluate the derivative f'(x) at different values of x within the interval [-2, 1]:
f'(-2) = 3(-2)^2 - 4(-2) + 1
= 12 + 8 + 1
= 21
f'(1) = 3(1)^2 - 4(1) + 1
= 3 - 4 + 1
= 0
The mean value theorem states that there exists a c in the interval (-2, 1) such that f'(c) = (f(1) - f(-2))/(1 - (-2)) = (1 - (-17))/(1 - (-2)) = 18/3 = 6.
Therefore, c = 6 satisfies the mean value theorem for the function f(x) = x^3 - 2x^2 + x + 1 in the interval [-2, 1].