Question

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval if `f(x) = 2x^2 − 3x + 1` and [0, 2]?

Answer 1

The Mean Value Theorem says that
if a function `f(x`) is differentiable over a range `[x_1,x_2]`
then there exists a value `c` within the range `[x_1,x_2]` such that

`f'(c) = (f(x_2) - (f(x_1)))/(x_2-x_1)`
(The right hand side of this is often referred to as the secant).

For the given function
`f(x) = 2x^2-3x+1` using `x_1=0` and `x_2=2`

`(f(2) - f(0))/(2-0)`

`=3/2`

`f'(x) = 4x-3`

If we (temporarily) assume the Mean Value Theorem holds then there must be a value `c` such that
`4c - 3 = 3/2`
which solves as
`c=9/8`

Testing this back in the equation of the Mean Value Theorem demonstrates that the theorem holds for the given equation.