Find c in the interval [-2,1] which satisfies the mean value theorem for f(x)=x^3+x+1
To satisfy the mean value theorem, we first need to check if the function f(x) is continuous on the interval [-2, 1] and differentiable on the open interval (-2, 1).
The function f(x) = x^3 + x + 1 is a polynomial function. Polynomials are continuous and differentiable for all real numbers. Therefore, f(x) is continuous on the interval [-2, 1] and differentiable on the open interval (-2, 1).
Now, let's find the average rate of change of f(x) over the interval [-2, 1] using the mean value theorem.
The average rate of change of f(x) over the interval [-2, 1] is given by:
f'(c) = (f(1) - f(-2))/(1 - (-2))
To find c, we need to find the derivative of f(x) and evaluate it at some point in the open interval (-2, 1).
Taking the derivative of f(x), we get:
f'(x) = 3x^2 + 1
Now, we need to evaluate f'(x) at c:
3c^2 + 1 = (f(1) - f(-2))/3
We can substitute the x-values of 1 and -2 into f(x):
3c^2 + 1 = ((1)^3 + (1) + 1 - ((-2)^3 + (-2) + 1))/3
3c^2 + 1 = (1 + 1 + 1 - (-8 - 2 + 1))/3
3c^2 + 1 = (3 - 5)/3
3c^2 + 1 = -2/3
Subtracting 1 from both sides:
3c^2 = -2/3 - 1
3c^2 = -2/3 - 3/3
3c^2 = -5/3
c^2 = -5/9
Since c^2 cannot be negative for real numbers, there is no value of c in the interval [-2, 1] that satisfies the mean value theorem for the function f(x) = x^3 + x + 1.
The mean value theorem states that for a function f(x) that is continuous on the interval [a, b] and differentiable on the open interval (a, b), there exists at least one value c in (a, b) such that the instantaneous rate of change at c is equal to the average rate of change over the interval [a, b].
In this case, we have f(x) = x^3 + x + 1, and we need to find the value of c within the interval [-2, 1] that satisfies the mean value theorem.
First, let's find the average rate of change of f(x) over the interval [-2, 1]. The average rate of change is given by the formula:
(Average Rate of Change) = [f(b) - f(a)] / (b - a)
where a and b are the endpoints of the interval. Plugging in the values for f(b), f(a), b, and a, we have:
(Average Rate of Change) = [f(1) - f(-2)] / (1 - (-2))
= [1^3 + 1 + 1 - (-2)^3 - (-2) - 1] / (3)
= [1 + 1 + 1 - (-8) + 2 - 1] / (3)
= [4] / (3)
= 4/3
Now, to find the value c that satisfies the mean value theorem, we need to find a value of c in the interval (-2, 1) such that the instantaneous rate of change of f(x) at c is equal to the average rate of change over the interval [-2, 1].
The derivative of f(x) is given by:
f'(x) = 3x^2 + 1
To find the value of c, we need to solve the equation f'(c) = 4/3.
Setting f'(c) equal to 4/3, we have:
3c^2 + 1 = 4/3
Multiply both sides of the equation by 3 to get rid of the fraction:
9c^2 + 3 = 4
Subtracting 3 from both sides:
9c^2 = 1
Dividing both sides by 9:
c^2 = 1/9
Taking the square root of both sides:
c = ±(1/3)
Since c must be in the interval (-2, 1), the only value that satisfies the mean value theorem is c = 1/3.
Therefore, the value of c that satisfies the mean value theorem for f(x) = x^3 + x + 1 in the interval [-2, 1] is c = 1/3.