(a) Given [tex]f(x) = x^3[/tex], the derivative is
[tex]f'(x)=3x^2[/tex]
which is exists for all [tex]x[/tex] in the domain of [tex]f[/tex], so [tex]]f[/tex] is differentiable everywhere and satisfies the mean value theorem. There is some number [tex]c[/tex] in the open interval (0, 2) such that
[tex]f'(c) = \dfrac{f(2) - f(0)}{2-0} \iff 3c^2 = \dfrac{8-0}2 = 4[/tex]
Solve for [tex]c[/tex] :
[tex]3c^2 = 4 \implies c^2 = \dfrac43 \implies \boxed{c = \dfrac2{\sqrt3}}[/tex]
We omit the negative square root since it doesn't belong to (0, 2). Graphically, the MVT tells us the tangent line to the curve [tex]f(x)=x^3[/tex] at [tex]x=\frac2{\sqrt3}[/tex] is parallel to the secant line through the endpoints of the given interval.
(b) [tex]f(x)=1+x+x^2[/tex] has derivative
[tex]f'(x)=1+2x[/tex]
By the MVT,
[tex]f'(c) = \dfrac{f(2)-f(0)}{2-0} \iff 1+2c = \dfrac{7-1}2 \implies \boxed{c = 1}[/tex]
(c) [tex]f(x) = \cos(2\pi x)[/tex] has derivative
[tex]f'(x) = -2\pi \sin(2\pi x)[/tex]
By the MVT,
[tex]f'(c) = \dfrac{f(2)-f(0)}{2-0} \iff -2\pi \sin(2\pi c) = \dfrac{0-0}2 \implies \sin(2\pi c) = 0 \\\\ \implies 2\pi c = n\pi \implies c = \dfrac n2[/tex]
where [tex]n[/tex] is any integer. There are 3 solutions in the interval (0, 2),
[tex]\boxed{c = \dfrac12, c = 1, c = \dfrac32}[/tex]
(Pictured is the situation with [tex]c=\frac12[/tex])
