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1).Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = 5x2 − 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.

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 =

2).Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval?
f(x) = ln x, [1, 7]
Yes, it does not matter if f is continuous or differentiable, every function satisfies the Mean Value Theorem.
Yes, f is continuous on [1, 7] and differentiable on (1, 7).
No, f is not continuous on [1, 7].
No, f is continuous on [1, 7] but not differentiable on (1, 7).
There is not enough information to verify if this function satisfies 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 satisfy the hypotheses, enter DNE).
c=

For the first question, to determine if the function satisfies the hypotheses of the Mean Value Theorem, we need to check if the function is continuous on the interval [0, 2] and differentiable on the open interval (0, 2).

The function f(x) = 5x^2 - 3x + 2 is a polynomial, which means it is continuous and differentiable on the domain of all real numbers.

Therefore, the function satisfies the hypotheses of the Mean Value Theorem.

To find all the numbers c that satisfy the conclusion of the Mean Value Theorem, we can use the formula:

c = (f(b) - f(a))/(b - a)

where a and b are the endpoints of the interval.

In this case, a = 0 and b = 2. Plugging in the values:

c = (f(2) - f(0))/(2 - 0)

Now, let's calculate f(2) and f(0):

f(2) = 5(2)^2 - 3(2) + 2 = 20 - 6 + 2 = 16

f(0) = 5(0)^2 - 3(0) + 2 = 0 - 0 + 2 = 2

Substituting the values:

c = (16 - 2)/(2 - 0) = 14/2 = 7

Therefore, the number c that satisfies the conclusion of the Mean Value Theorem is c = 7.

For the second question, the function f(x) = ln x is continuous and differentiable on the interval (1, 7). This means it satisfies the hypotheses of the Mean Value Theorem.

To find the numbers c that satisfy the conclusion of the Mean Value Theorem, we can again use the formula:

c = (f(b) - f(a))/(b - a)

In this case, a = 1 and b = 7. Plugging in the values:

c = (f(7) - f(1))/(7 - 1)

Now, let's calculate f(7) and f(1):

f(7) = ln(7)

f(1) = ln(1) = 0

Substituting the values:

c = (ln(7) - 0)/(7 - 1) = ln(7)/6

Therefore, the number c that satisfies the conclusion of the Mean Value Theorem is c = ln(7)/6.

To determine whether a function satisfies the hypotheses of the Mean Value Theorem on a given interval, there are two conditions that need to be met:

1. The function must be continuous on the closed interval [a, b].
2. The function must be differentiable on the open interval (a, b).

Let's analyze the given functions:

1) f(x) = 5x^2 - 3x + 2, [0, 2]
- The function is a polynomial, which makes it continuous and differentiable on the entire real line.
- Therefore, f(x) is continuous on the closed interval [0, 2].
- The derivative of f(x) is 10x - 3, which is defined and continuous for all real numbers.
- Hence, f(x) is differentiable on the open interval (0, 2).

Therefore, f(x) satisfies the hypotheses of the Mean Value Theorem on the given interval.

To find all the numbers c that satisfy the conclusion of the Mean Value Theorem, we need to find the value of c within the open interval (0, 2) where:

f'(c) = (f(2) - f(0))/(2-0)

Substituting the values from the function:
10c - 3 = (5(2)^2 - 3(2) + 2 - (5(0)^2 - 3(0) + 2))/(2-0)

Simplifying the equation:
10c - 3 = (20 - 6 + 2)/(2)
10c - 3 = 16/2
10c - 3 = 8
10c = 11
c = 11/10

Therefore, c = 11/10 is the number that satisfies the conclusion of the Mean Value Theorem for the given function on the interval [0, 2].

2) f(x) = ln x, [1, 7]
- The natural logarithm function, ln x, is continuous and differentiable on its domain, which includes (1, 7) and [1, 7].
- Hence, f(x) is continuous on the closed interval [1, 7], and differentiable on the open interval (1, 7).

Therefore, f(x) satisfies the hypotheses of the Mean Value Theorem on the given interval.

To find all the numbers c that satisfy the conclusion of the Mean Value Theorem, we need to find the value of c within the open interval (1, 7) where:

f'(c) = (f(7) - f(1))/(7-1)

Using the properties of the natural logarithm:
1/c = (ln(7) - ln(1))/(6)
1/c = ln(7)/6

Simplifying the equation:
c = 6/ln(7)

Therefore, c = 6/ln(7) is the number that satisfies the conclusion of the Mean Value Theorem for the given function on the interval [1, 7].

#1 The MVT requires that f(x) be continuous and differentiable. Luckily, all polynomials meet these requirements.

f(x) = 5x^2-3x+2
f'(x) = 10x-3
f(0) = 2
f(2) = 16
So, we want c where f'(c) = 7
10x-3 = 6
x = 0.9
The tangent line at x=0.9 is
y-3.35 = 7(x-.9)
y = 7x-2.95

http://www.wolframalpha.com/input/?i=plot+y%3D5x^2-3x%2B2%2C+y%3D7x-2.95

#2
lnx satisfies the MVT on [1,7]
f(1) = 0
f(7) = ln7
so the slope of the secant is ln7 / 6
so, we want f'(x) = 1/x = ln7/6
c = 6/ln7 = 3.083, which is in [1,7]
f(6/ln7) = 1.126
So, the tangent line is
y-1.126 = (ln7/6)(x-3.083)

http://www.wolframalpha.com/input/?i=plot+y%3Dln+x+%2C+y%3D+%28ln7%2F6%29%28x-3.083%29%2B1.126+for+x%3D1..7