Question

A. Given f(x)=x3 use the Mean Value Theorem and find all point c, 0

Answer 1

a)
`x \approx 1.155`, b)
`x = \pm(i)/(2)`, for all
`i = \{0,1,2,3,... \}`

Explanation:

a) The slope associated with the mean value is:

`f'(c) =(f(2)-f(0))/(2 - 0)`

`f'(c) = (8-0)/(2-0)`

`f'(c) = 4`

Let differentiate the function:

`f'(x) = 3\cdot x^(2)`

The value associated with the slope is:

`3\cdot x^(2) = 4`

`x = \sqrt{(4)/(3) }`

`x \approx 1.155`

b) The slope associated with the mean value is:

`f'(c) =(f(2)-f(0))/(2 - 0)`

`f'(c) = (1-1)/(2-0)`

`f'(c) = 0`

Let differentiate the function:

`f'(x) = - 2\pi \cdot \sin (2\pi\cdot x)`

The value associated with the slope is:

`-2\pi\cdot \sin (2\pi\cdot x) = 0`

`\sin (2\pi\cdot x) = 0`

`2\pi\cdot x = \sin^(-1) 0`

`2\pi \cdot x = \pm \pi \cdot i`, for all
`i = \{0,1,2,3,... \}`

`x = \pm(i)/(2)`, for all
`i = \{0,1,2,3,... \}`