Answer:
a) [tex]x \approx 1.155[/tex], b) [tex]x = \pm\frac{i}{2}[/tex], for all [tex]i = \{0,1,2,3,... \}[/tex]
Step-by-step explanation:
a) The slope associated with the mean value is:
[tex]f'(c) =\frac{f(2)-f(0)}{2 - 0}[/tex]
[tex]f'(c) = \frac{8-0}{2-0}[/tex]
[tex]f'(c) = 4[/tex]
Let differentiate the function:
[tex]f'(x) = 3\cdot x^{2}[/tex]
The value associated with the slope is:
[tex]3\cdot x^{2} = 4[/tex]
[tex]x = \sqrt{\frac{4}{3} }[/tex]
[tex]x \approx 1.155[/tex]
b) The slope associated with the mean value is:
[tex]f'(c) =\frac{f(2)-f(0)}{2 - 0}[/tex]
[tex]f'(c) = \frac{1-1}{2-0}[/tex]
[tex]f'(c) = 0[/tex]
Let differentiate the function:
[tex]f'(x) = - 2\pi \cdot \sin (2\pi\cdot x)[/tex]
The value associated with the slope is:
[tex]-2\pi\cdot \sin (2\pi\cdot x) = 0[/tex]
[tex]\sin (2\pi\cdot x) = 0[/tex]
[tex]2\pi\cdot x = \sin^{-1} 0[/tex]
[tex]2\pi \cdot x = \pm \pi \cdot i[/tex], for all [tex]i = \{0,1,2,3,... \}[/tex]
[tex]x = \pm\frac{i}{2}[/tex], for all [tex]i = \{0,1,2,3,... \}[/tex]
