The value of constant c between 1 and 9 such that the average value of the function f(x) on the interval [1,9] is equal to is 5.
<h3>What is mean value theorem?</h3>
Mean value theorem is the theorem which is used to find the behavior of a function.
The function given as,
![f(x) = 6x^2 - 4x + 2](https://tex.z-dn.net/?f=f%28x%29%20%3D%206x%5E2%20-%204x%20%2B%202)
The value of function at 0,
![f(0) = 6(0)^2 - 4(0) + 2\\f(0)=2](https://tex.z-dn.net/?f=f%280%29%20%3D%206%280%29%5E2%20-%204%280%29%20%2B%202%5C%5Cf%280%29%3D2)
Differentiate the given equation,
![f(x)' = 6\times2x - 4\times1 + 0\\f(x)' = 12x - 4\\f(x)' = 12x -4](https://tex.z-dn.net/?f=f%28x%29%27%20%3D%206%5Ctimes2x%20-%204%5Ctimes1%20%2B%200%5C%5Cf%28x%29%27%20%3D%2012x%20-%204%5C%5Cf%28x%29%27%20%3D%2012x%20-4)
If the constant c between 1 and 9 such that the average value of the function f(x) on the interval (1,9], then,
![f(c)'=12c-4](https://tex.z-dn.net/?f=f%28c%29%27%3D12c-4)
Using Lagrange's mean value theorem,
![f(c)'=\dfrac{f(9)-f(1)}{9-1}\\12c-4=\dfrac{452-4}{9-1}\\12c=54+4\\c=\dfrac{60}{12}\\c=5](https://tex.z-dn.net/?f=f%28c%29%27%3D%5Cdfrac%7Bf%289%29-f%281%29%7D%7B9-1%7D%5C%5C12c-4%3D%5Cdfrac%7B452-4%7D%7B9-1%7D%5C%5C12c%3D54%2B4%5C%5Cc%3D%5Cdfrac%7B60%7D%7B12%7D%5C%5Cc%3D5)
Thus, the value of constant c between 1 and 9 such that the average value of the function f(x) on the interval [1,9] is equal to is 5.
Learn more about the mean value theorem here;
brainly.com/question/15115079
#SPJ1