Does the function satisfy the hypotheses of the mean value theorem on the given interval? f(x) = 3x2 − 4x + 1, [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. 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 =

The true statement is: (b) Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable

Mean value theorem states that:

If [a,b] and

on (a,b),

Then there is a point c in (a,b), such that:

The function is given as:

And the interval is:

We have

This becomes

Calculate f(2) and f(0)

So, we have:

Recall that:

Differentiate

Substitute c for x

Substitute 2 for f'(c)

Collect like terms

Divide both sides by 6

The interval is given as: [0,2]

The value of c is true for interval (0,2).

Hence, the true statement is:

(b) Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable

Read more about mean value theorems at:

link

is a polynomial and thus continuous everywhere and differentiable on any open interval. (second option)

The MVT then guarantees the existence of such that

We have

so

80 the answer is 80

step-by-step explanation:

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic