Question

Determine if the Mean Value Theorem for Integrals applies to the function f(x) = 3 - x2 on the interval the closed interval from 0 to the square root of 3 . If so, find the x-coordinates of the point(s) guaranteed by the theorem

Answer 1

Answer: Yes, Mean Value Theorem for Integrals applies for the function
`f(x)=3-x^(2)` and values of x are 1 and -1

Explanation: Mean Value Theorem for Integrals states that if a function is continuous on a closed interval, there is a value c on the interval such that

`f(c)=(1)/(b-a) \int\limits^b_a {f(x)} \, dx`

where a and b are the closed interval given

As the graph of
`f(x)=3-x^(2)` shown below, the function is continuous on the interval [0,
`√(3)`], so the theorem applies.

To find the x-coordinates, first determine value of f(c) at [0,
`√(3)`]:

`f(c)=(1)/(√(3) -0) \int\limits {3-\sqrt{x^(2)} } \, dx`

`f(c)=(1)/(√(3))[3x-(x^(3))/(3) ]`

`f(c)=(1)/(√(3))[ 3√(3)-((√(3))^(3))/(3) ]`

`f(c)=(1)/(√(3) ) [2√(3) ]`

f(c) = 2

The x-coordinates will be:

`f(x)=3-x^(2)`

`2=3-x^(2)`

`x^(2)=1`

`x=√(1)`

x = ±1

The values for x guaranteed by the Mean Value Theorem are +1 and -1.