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 =

Answer:a) Yes, f is continuous on [0, 2] and differentiable on (0, 2) since polynomials are continuous and differentiable on Rb) c=1Step-by-step explanation:a)The given function is;[tex]f(x)=4x^2-3x+2[/tex]For this function to satisfy the Mean Value Theorem, it must;i) be continuous on [0,2]ii)be differentiable on (0,2)Since [tex]f(x)=4x^2-3x+2[/tex] is a polynomial function, it is continuous on [0,2] and differentiable on (0,2) because polynomial functions are continuous and differentiable on the entire real numbers.b)Since the above two hypotheses are satisfied,it implies that, there is a certain [tex]c\in (0,2)[/tex] such that; [tex]f'(c)=\frac{f(2)-f(0)}{2-0}[/tex][tex]f'(x)=8x-3[/tex][tex]8c-3=\frac{12-2}{2}[/tex][tex]8c-3=\frac{10}{2}[/tex][tex]8c-3=5[/tex][tex]8c=5+3[/tex][tex]8c=8[/tex][tex]c=1[/tex]Therefore the number that satisfies the conclusion of the Mean Value Theorem is c=1.We can see from the graph that at x=1, the secant line is parallel to the tangent on the interval [0,2]