Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 4x2 − 3x + 2, [0, 2] yes, it does not matter if f is continuous or differentiable, every function satifies the mean value theorem. yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on double-struck r. no, f is not continuous on [0, 2]. no, f is continuous on [0, 2] but not differentiable on (0, 2). there is not enough information to verify if this function satifies the mean value theorem. correct: your answer is correct. if it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the mean value theorem. (enter your answers as a comma-separated list. if it does not satisify the hypotheses, enter dne). c =

c = 1

Step-by-step explanation:

The formula for the MVT is

where f'(c) is the derivative of the function:  8x - 3;

f(b) is the function evaluated at an x value of 2 (the second number in the interval):  f(b) = 12;

f(a) is the function evaluated at an x value of 1 (the first number in the interval:  f(a) = 2

Setting up:

So basically what we end up with is

8c - 3 = 5 and

8c = 8 so

c = 1

This function does in fact satisfy the requirements for the MVT:  it is continuous on the interval and it is differentiable on the interval as polynomials by nature are continuous and differentiable on all values of x.

step-by-step explanation: a reaction is first order. if the initial reactant concentration is 0.0200 m, and 25.0 days later the concentration is 6.25 x 10-4 m, then its half-life is:

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.