MATH SOLVE

4 months ago

Q:
# Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? f(x) = 3x2 β 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 =

Accepted Solution

A:

[tex]f(x)=3x^2-4x+1[/tex] is a polynomial and thus continuous everywhere and differentiable on any open interval. (second option)The MVT then guarantees the existence of [tex]c\in(0,2)[/tex] such that[tex]f'(c)=\dfrac{f(2)-f(0)}{2-0}=\dfrac{5-1}2=2[/tex]We have[tex]f'(x)=6x-4[/tex]so[tex]6c-4=2\implies6c=6\implies c=1[/tex]