1.) The absolute maximum of the functionf(x)=1+x+x2x on the interval [-2, 2] is 1/2 at x = 1, and the absolute minimum is -1/2 at x = -1.

For 2) and 3) the verification is in the explanation below.

Explanation:

1. The function given is:f(x)=1+x+x2x

2. To find the absolute maximum and minimum of this function on the interval [2,2], we need to follow these steps:

a) Find the critical points of the function within the interval:

• Take the derivative of the function:f(x)=(1+x+x2)21x2
• Set the derivative equal to 0 and solve for x:1x2=0x=1
• The critical points are x = 1 and x = -1, as they are the only points where the derivative is 0 within the interval [-2, 2].

b) Evaluate the function at the critical points and the endpoints of the interval:

• f(1)=1+1+11=31
• f(1)=11+11=31
• f(2)=12+42=32
• f(2)=1+2+42=72

3. Comparing the function values at the critical points and the endpoints, we can determine the absolute maximum and minimum:

• The absolute maximum is 1/2, which occurs at x = 1.
• The absolute minimum is -1/2, which occurs at x = -1.

Therefore, the absolute maximum of the functionf(x)=1+x+x2x on the interval [-2, 2] is 1/2 at x = 1, and the absolute minimum is -1/2 at x = -1.

2.) To verify thatf(x) satisfies the hypotheses of Rolle's Theorem:

1. Continuity:

The functionf(x)=sin(2x) is continuous on the closed interval [0,]. This is because the sine function is continuous everywhere, and the composition of continuous functions is also continuous.

2. Differentiability:

The functionf(x)=sin(2x) is differentiable on the open interval (0,). This is because the sine function is differentiable everywhere, and the derivative ofsin(2x) is f(x)=2cos(2x), which is defined on (0,).

3. Endpoint conditions: We havef(0)=sin(0)=0 and f()=sin(2)=0, so the function satisfies the condition f(0)=f(). Therefore, the function f(x)=sin(2x)satisfies all the hypotheses of Rolle's Theorem on the interval [0,].

To find the pointsc that satisfy the conclusion of Rolle's Theorem, we need to find the critical points of the function, i.e., the points where the derivative is equal to 0. The derivative off(x)=sin(2x) is f(x)=2cos(2x).

To find the pointsc where f(c)=0, we set2cos(2c)=0 and solve for c. The solutions are c=4and c=43, as these are the values ofc where cos(2c)=0.

3.) To verify thatf(x) satisfies the hypotheses of the Mean Value Theorem:

1. Continuity:

f(x)=e3xis continuous on the closed interval[0,3] since the exponential function is continuous everywhere.

2. Differentiability:

f(x)=3e3x is defined on the open interval (0,3), sof(x) is differentiable on (0,3).

Therefore, f(x)=e3xsatisfies the hypotheses of the Mean Value Theorem on the interval [0,3].

To find the numbersc that satisfy the conclusion of the Mean Value Theorem:

The Mean Value Theorem states that there exists at least one numberc in the open interval(a,b) such that:

f(c)=baf(b)f(a)

In this case, a=0and b=3, so we have:

f(c)=30f(3)f(0)

Substituting the given function f(x)=e3x,

we get:3e3c=3e9e0

Simplifying, we get:3e3c=3e91

Solving forc we get:c=1

Therefore, the numberc that satisfies the conclusion of the Mean Value Theorem is c=1.

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