does the function satisfy the hypotheses of the mean value theorem on the given interval?\n$f(x)=4x^{2}-3x +…

does the function satisfy the hypotheses of the mean value theorem on the given interval?\n$f(x)=4x^{2}-3x + 2$, $0,2$\no yes, it does not matter if $f$ is continuous or differentiable, every function satisfies the mean value theorem.\no yes, $f$ is continuous on $0,2$ and differentiable on $(0,2)$ since polynomials are continuous and differentiable on $mathbb{r}$.\no no, $f$ is not continuous on $0,2$.\no no, $f$ is continuous on $0,2$ but not differentiable on $(0,2)$.\no there is not enough information to verify if this function satisfies the mean value theorem.\nif 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 satisfy the hypotheses, enter dne)\n$c=$

does the function satisfy the hypotheses of the mean value theorem on the given interval?\n$f(x)=4x^{2}-3x + 2$, $0,2$\no yes, it does not matter if $f$ is continuous or differentiable, every function satisfies the mean value theorem.\no yes, $f$ is continuous on $0,2$ and differentiable on $(0,2)$ since polynomials are continuous and differentiable on $mathbb{r}$.\no no, $f$ is not continuous on $0,2$.\no no, $f$ is continuous on $0,2$ but not differentiable on $(0,2)$.\no there is not enough information to verify if this function satisfies the mean value theorem.\nif 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 satisfy the hypotheses, enter dne)\n$c=$

Answer

Explanation:

Step1: Check continuity and differentiability

Since (f(x)=4x^{2}-3x + 2) is a polynomial function. Polynomial functions are continuous on (\mathbb{R}), so (f(x)) is continuous on ([0,2]). Polynomial functions are differentiable on (\mathbb{R}), so (f(x)) is differentiable on ((0,2)).

Step2: Apply the Mean Value Theorem formula

The Mean Value Theorem states that (f^{\prime}(c)=\frac{f(b)-f(a)}{b - a}), where (a = 0), (b=2). First, find (f(0)) and (f(2)): (f(0)=4\times0^{2}-3\times0 + 2=2) (f(2)=4\times2^{2}-3\times2 + 2=16-6 + 2=12) Then (\frac{f(2)-f(0)}{2-0}=\frac{12 - 2}{2}=5) Next, find the derivative (f^{\prime}(x)): Using the power rule ((x^{n})^{\prime}=nx^{n - 1}), (f^{\prime}(x)=8x-3) Set (f^{\prime}(c)=8c-3) equal to (5) (from (\frac{f(2)-f(0)}{2-0})): (8c-3 = 5)

Step3: Solve for (c)

Add (3) to both sides of the equation (8c-3 = 5): (8c=5 + 3=8) Divide both sides by (8): (c = 1)

Answer:

Yes, (f) is continuous on ([0,2]) and differentiable on ((0,2)) since polynomials are continuous and differentiable on (\mathbb{R}). (c = 1)