Let f(x) be a real-valued function defined on the interval [0, 1] and satisfying the following conditions: f(0) = 0 f(1) = 1 f'(x) exists for all x in (0, 1) If the function f(x) satisfies the Mean Value Theorem on the interval (0, 1), then there exists a point c in (0, 1) such that: a) f'(c) = 1 b) f'(c) = 0 c) f'(c) = 2 d) f'(c) = -1 Choose the correct option and provide a brief explanation for your choice.

1) Consider f(x) = 3x4 -8x³ +6 a) Find all critical points of the function (make sure to consider if the derivative is defined everywhere or not!). Label them in increasing order, 21, 22, 23, etc (for how many you find). b) Use the first derivative test to determine if x₁ is a local max, min, or neither. Find the y value associated with this a value and write it as a point (x, y). 1 c) Use the second derivative test to determine if x2 is a local max, min, or neither. Find the y value associated with this a value and write it as a point (x, y).

Consider the function f(x) = on the interval [2, 12]. (A) Find the average or mean slope of the function on this interval, i.e. f(12) – f(2) 12 – 2 (B) By the Mean Value Theorem, we know there exists a c in the open interval (2, 12) such that f'(c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below. List of values:

For f(x) = 1+x+x^2, use the mean value theorem to show that there is real number c, 0 < c < 2, such that f(2)-f(0) = f'(c)(2-0). Find all such c.