Question

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

Answer 1

Answer:

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

Step-by-step explanation:

a) The slope associated with the mean value is:

$f'(c) =\frac{f(2)-f(0)}{2 - 0}$

$f'(c) = \frac{8-0}{2-0}$

$f'(c) = 4$

Let differentiate the function:

$f'(x) = 3\cdot x^{2}$

The value associated with the slope is:

$3\cdot x^{2} = 4$

$x = \sqrt{\frac{4}{3} }$

$x \approx 1.155$

b) The slope associated with the mean value is:

$f'(c) =\frac{f(2)-f(0)}{2 - 0}$

$f'(c) = \frac{1-1}{2-0}$

$f'(c) = 0$

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 $i = \{0,1,2,3,... \}$

$x = \pm\frac{i}{2}$, for all $i = \{0,1,2,3,... \}$