Question

Let f(x) = 6x^2 - 4x + 2. Find a constant c between 1 and 9 such that the average value of the function f(x) on the interval (1,9] is equal to fc).
A. 4.8735
B. 5.5402
C. 5.5721
D. -4.8735
E. 164​

Answer 1

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.

What is mean value theorem?

Mean value theorem is the theorem which is used to find the behavior of a function.

The function given as,

$f(x) = 6x^2 - 4x + 2$

The value of function at 0,

$f(0) = 6(0)^2 - 4(0) + 2\\f(0)=2$

Differentiate the given equation,

$f(x)' = 6\times2x - 4\times1 + 0\\f(x)' = 12x - 4\\f(x)' = 12x -4$

If the constant c between 1 and 9 such that the average value of the function f(x) on the interval (1,9], then,

$f(c)'=12c-4$

Using Lagrange's mean value theorem,

$f(c)'=\dfrac{f(9)-f(1)}{9-1}\\12c-4=\dfrac{452-4}{9-1}\\12c=54+4\\c=\dfrac{60}{12}\\c=5$

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;

https://brainly.com/question/15115079

#SPJ1