Question

Calculus. Please answer question attached.

Answer 1

(a) Given
`f(x) = x^3`, the derivative is

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

which is exists for all
`x` in the domain of
`f`, so
`]f` is differentiable everywhere and satisfies the mean value theorem. There is some number
`c` in the open interval (0, 2) such that

`f'(c) = (f(2) - f(0))/(2-0) \iff 3c^2 = \frac{8-0}2 = 4`

Solve for
`c` :

`3c^2 = 4 \implies c^2 = \frac43 \implies \boxed{c = \frac2{\sqrt3}}`

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
`f(x)=x^3` at
`x=\frac2{\sqrt3}` is parallel to the secant line through the endpoints of the given interval.

(b)
`f(x)=1+x+x^2` has derivative

`f'(x)=1+2x`

By the MVT,

`f'(c) = (f(2)-f(0))/(2-0) \iff 1+2c = \frac{7-1}2 \implies \boxed{c = 1}`

(c)
`f(x) = \cos(2\pi x)` has derivative

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

By the MVT,

`f'(c) = (f(2)-f(0))/(2-0) \iff -2\pi \sin(2\pi c) = \frac{0-0}2 \implies \sin(2\pi c) = 0 \\\\ \implies 2\pi c = n\pi \implies c = \frac n2`

where
`n` is any integer. There are 3 solutions in the interval (0, 2),

`\boxed{c = \frac12, c = 1, c = \frac32}`

(Pictured is the situation with
`c=\frac12`)