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1. Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?f(x) = eāˆ’2x, [0, 1]Yes, f is continuous and differentiable on double-struck R, so it is continuous on [0, 1] and differentiable on (0, 1) .There is not enough information to verify if this function satisfies the Mean Value Theorem. No, f is not continuous on [0, 1].Yes, it does not matter if f is continuous or differentiable; every function satisfies the Mean Value Theorem.No, f is continuous on [0, 1] but not differentiable on (0, 1).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 satisfy the hypotheses, enter DNE).c = ____ ?2. Verify that the function satisfies the three hypotheses of Rolle's Theorem on the given interval. Then find all numbers c that satisfy the conclusion of Rolle's Theorem. (Enter your answers as a comma-separated list.)f(x) = x3 āˆ’ x2 āˆ’ 6x + 9, [0, 3]c =

Respuesta :

LammettHash

1. [tex]f(x)=e^{-2x}[/tex] is continuous and differentiable everywhere, so it is continuous on [0, 1] and differentiable on (0, 1), and the MVT is satisfied.

We have

[tex]f'(x)=-2e^{-2x}[/tex]

and the MVT says there is some [tex]c\in(0,1)[/tex] such that

[tex]f'(c)=\dfrac{f(1)-f(0)}{1-0}[/tex]

We find that

[tex]-2e^{-2c}=e^{-2}-1\implies c=1+\ln\sqrt{\dfrac2{e^2-1}}[/tex]

2. Rolle's theorem has the same conditions as the mean value theorem, with the added condition that [tex]f(x)[/tex] attains the same values at both endpoints of the interval in question.

Here, [tex]f(x)[/tex] is a polynomial, so it's continuous and differentiable everywhere, and [tex]f(0)=9[/tex] and [tex]f(3)=9[/tex], so Rolle's theorem is satisfied.

It then says that there exists [tex]c\in(0,3)[/tex] such that

[tex]f'(c)=\dfrac{f(3)-f(0)}{3-0}=0[/tex]

The derivative is

[tex]f'(x)=3x^2-2x-6[/tex]

and we have

[tex]3c^2-2c-6=0\implies c=\dfrac{1\pm\sqrt{19}}3[/tex]

Take the solution with the positive square root (the other solution is negative) so that

[tex]c=\dfrac{1+\sqrt{19}}3[/tex]