Question

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 =

Answer 1

$f(x)=3x^2-4x+1$ is a polynomial and thus continuous everywhere and differentiable on any open interval. (second option)

The MVT then guarantees the existence of $c\in(0,2)$ such that

$f'(c)=\dfrac{f(2)-f(0)}{2-0}=\dfrac{5-1}2=2$

We have

$f'(x)=6x-4$

so

$6c-4=2\implies6c=6\implies c=1$

Answer 2

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 $\mathbf{f(x)\ is\ continuous}$ [a,b] and

$\mathbf{f(x)\ is\ differentiable}$ on (a,b),

Then there is a point c in (a,b), such that: $\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}$

The function is given as:

$\mathbf{f(x) = 3x^2 - 4x + 1}$

And the interval is: $\mathbf{[0,2]}$

We have

$\mathbf{f'(c) = \frac{f(b) - f(a)}{b - a}}$

This becomes

$\mathbf{f'(c) = \frac{f(2) - f(0)}{2 - 0}}$

$\mathbf{f'(c) = \frac{f(2) - f(0)}{2}}$

Calculate f(2) and f(0)

$\mathbf{f(2) = 3\times 2^2 - 4\times 2 + 1 = 5}$

$\mathbf{f(0) = 3\times 0^2 - 4\times 0 + 1 = 1}$

So, we have:

$\mathbf{f'(c) = \frac{5-1}{2}}$

$\mathbf{f'(c) = \frac{4}{2}}$

$\mathbf{f'(c) = 2}$

Recall that:

$\mathbf{f(x) = 3x^2 - 4x + 1}$

Differentiate

$\mathbf{f'(x)= 6x - 4}$

Substitute c for x

$\mathbf{f'(c)= 6c - 4}$

Substitute 2 for f'(c)

$\mathbf{ 6c - 4 = 2}$

Collect like terms

$\mathbf{ 6c = 4 + 2}$

$\mathbf{ 6c = 6}$

Divide both sides by 6

$\mathbf{c = 1}$

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:

https://brainly.com/question/3957181