Enter the function f(x), specify the interval [a,b], and click “Calculate” to find the point “c” that satisfies the mean value theorem.
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This online mean value theorem calculator finds the point "c" within the interval [a,b] for the given function f(x) when the function satisfies the condition of continuity on the closed interval [a,b] and differentiability over the open interval (a,b)
The mean value theorem is used to evaluate the behavior of a function. The mean value theorem asserts that if the “f” is a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), then there is at least one point "c" on the open interval (a, b). At this point, the instantaneous rate of change of function is equal to the average rate of change of that function.
The mean value theorem formula is:
\(\ f’ (c) =\frac{f(b) – f (a)}{b - a}\)
Where:
The mean value theorem for integral states that if a function f(x) is continuous, then a point “c” exists within the interval [a,b] such that:
f(c) = (1 / (b - a)) ∫ab f(x) dx
Using an online integral calculator helps you to evaluate the integrals of the functions for the variable involved.
Example:
Find the value of f (x)=11x^2 - 6x - 3 on the interval [4,8].
Solution:
In the given equation (f) is continuous on [4, 8].
\(\ F (C) = 1/b - a ∫ f(x) dx = 1/ 8 - 4∫_4^8 (11x^2 - 6x - 3) dx\)
\(\ = 1/4 [x^3 -x^2]^8_4\)
\(\ = 1/4 [(216 - 36) - (8 - 4)]\)
\(\ = 1/4 [(180 - 4)]\)
\(\ =\frac{176}{4} =\ 44\)
Here the value of c is 44 which provides the average value of the given function.
Now put x=16 in the function.
\(\ f(x)=11x^2\ -\ 6x\ -\ 3 =\ 44\)
\(\ =11x^2\ -\ 6x\ -\ 47\)
\(\ =(x\ +\ 2.32)(x\ -\ 2.80)\ =\ 0\)
Hence 2.80 is the value of c. You can get the same detailed result from the online mean value theorem calculator by plugging in similar values and intervals.
Cauchy’s mean value theorem is the generalization of the mean value theorem. It states: if the function “g” and “f” both are continuous on the end interval [a, b] and differentiable on the start interval (a, b), then there exists point “c” e(a, b), such that:
\(\text{(f(b) - f (a)) g’c = (g(b) - g (a)) f’c}\)
Here’s g (a) ≠ g (b) and g’ (c) ≠0, so this is equivalent to:
\(\text{f’(c) / g’(c) =[f(b) - f(a)]/[g(b) - g(a)] }\)
Using an online derivative calculator, you can quickly find the derivative of the function with respect to a given variable.
Example:
Solution:
f(x) is a polynomial function and is differentiable for all real numbers. Let evalute f(x) at x = -4 and x = 2
\(\ f(-4) =\ -4(-4)^3\ +\ 6(-4) - 2 = 20\)
\(\ f(2) =\ -4(2)^3\ +\ 6(2) - 2 = - 4\)
Now, substitute the values in [f(b) - f(a)] / (b - a)
\(\frac{f(b) - f(a)}{b - a} =\frac{-6 - 4}{2 - (-4)} = -2\)
Let us now find f '(x)
\(\ f '(x) =\ - 6x^2 + 6\)
We now create an equation, which is based on f '(c) = [f(b) - f(a)] / (b - a)
\(\ -6c + 6 = -2\)
You can find the value of c by using the mean value theorem calculator:
\(\ c = 2 \sqrt{(1/3)}\ and\ c =\ - 2 \sqrt{(1/3)}\)
Rolle’s theorem is the special case of the mean value theorem. In this case, the mean value theorem is applied but with the additional condition which is:
Example:
Find all values of point c in the interval [−4,0]such that f′(c)=0.Where f(x)=x^2+2x.
Solution:
First of all, check the function f(x) that satisfies all the states of Rolle’s theorem.
\(\ f(−2)=(−4)2+2⋅(−4)=\ 0\)
\(\ f(0)=02+2⋅0=\ 0\)
\(\ f(−4)=\ f(0)\)
We can also use Rolle’s theorem calculator to find the point c
\(\ f′(x)=(x2+2x)′=\ 2x\ +\ 2\)
Now, solve the equation f′(c)=0:
\(\ f′(c)\ =\ 2c\ +\ 2\ =\ 0\)
\(\ c\ =\ −1\)
Thus,
\(\ f′(c)\ =\ 0\ for\ c\ =\ −1\)
Follow these steps:
This theorem also known as the First Mean Value Theorem allows showing the increment of a given function (f) on a specific interval through the value of a derivative at an intermediate point.
References:
From the source of Wikipedia: Cauchy's mean value theorem, Proof of Cauchy's mean value theorem, Mean value theorem in several variables.
From the source of Pauls Online Notes: The Mean Value Theorem, Rolle’s Theorem, Proofs From Derivative Applications.
From the source of Calc Workshop: Mean Value Theorem for Integrals, Average Value, Mean Value Theorem.
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