Hi I am having some trouble with these few quetions I would appreciate some help so that I can understand them better.


1) What, if anything, does the mean value theorem guarantee for the given function on this interval?

a) f(x) = x^2 - 2x + 5 on [1,4]

--I am a bit uncertain on how to answer this, I started out with

f '(c) = f(b) - f(a) / b-a = 13 - 4 / 3 = 9 / 3 = 3.

Then I plugged in 3 for f(x) and got 8.

Does this mean that (3,8) is a critical point?

b) g(x) = 8 / [(x-2)^2] on [1,4]

I am sure this one needs a similiar approach to the last


Lastly:

What values c (if any) are predictable by the mean value theorem for the function f(x) = (x-2)^3 on the interval [0,2]?

I proceeded similiarly here like the last question.

f '(c) turned out to be 4, and f(4) was 8.

I would greatly appreciate some help, since I am having trouble understanding the question and what it is asking.

Thanks!

f '(c) = f(b) - f(a) / b-a = 13 - 4 / 3 = 9 / 3 = 3.

There exists a point c in the interval [1,4] such that f'(c) = 3

ok thanks

4+20=46

verify that the function to satisfies roller's theorem of f /x/=x-3x+2x+5 [0,2].

Sure, I'd be happy to help you understand these questions better.

1a) The mean value theorem states that for a differentiable function on a closed interval, there exists at least one point in the interval where the instantaneous rate of change (the derivative) is equal to the average rate of change over the interval. In this case, the function is f(x) = x^2 - 2x + 5 on the interval [1,4].

To use the mean value theorem, we need to find the derivative of f(x). So, let's find f '(x):

f '(x) = d/dx (x^2 - 2x + 5)
= 2x - 2

Now, let's find the average rate of change of f(x) over the interval [1,4]:

Average rate of change = (f(4) - f(1))/(4 - 1)
= (17 - 4)/(3)
= 13/3

According to the mean value theorem, there exists at least one point c in the interval [1,4] where f '(c) = average rate of change. In other words, there exists a point c in the interval [1,4] such that f '(c) = 13/3.

To determine if (3,8) is a critical point, we need to check if f '(3) = 13/3. Let's substitute x = 3 into f '(x):

f '(3) = 2(3) - 2
= 4

Since f '(3) ≠ 13/3, (3,8) is not a critical point. The mean value theorem does not guarantee any specific critical points for the given function on the interval [1,4].

1b) Let's apply the same process to the function g(x) = 8/[(x-2)^2] on the interval [1,4].

First, find g '(x):

g '(x) = d/dx (8/[(x-2)^2])
= -16/(x-2)^3

Next, find the average rate of change of g(x) over the interval [1,4]:

Average rate of change = (g(4) - g(1))/(4 - 1)
= (2 - 8)/(3)
= -6/3
= -2

According to the mean value theorem, there exists at least one point c in the interval [1,4] where g '(c) = average rate of change. In other words, there exists a point c in the interval [1,4] such that g '(c) = -2.

To determine if any critical points exist, we need to solve g '(x) = -2. Let's set the derivative equal to -2 and solve for x:

-16/(x-2)^3 = -2
Multiply both sides by (x-2)^3:
-16 = -2(x-2)^3
Divide both sides by -2:
8 = (x-2)^3

Taking the cube root of both sides:
2 = x-2
Add 2 to both sides:
x = 4

Therefore, the only value of c that satisfies g '(c) = -2 is c = 4. The mean value theorem predicts that at x = 4, g(x) has an instantaneous rate of change equal to the average rate of change over the interval [1,4].

Lastly, for the function f(x) = (x-2)^3 on the interval [0,2], let's apply the same steps:

Find f '(x):
f '(x) = d/dx ((x-2)^3)
= 3(x-2)^2

Find the average rate of change of f(x) over the interval [0,2]:

Average rate of change = (f(2) - f(0))/(2 - 0)
= (0 - (-8))/(2)
= 8/2
= 4

According to the mean value theorem, there exists at least one point c in the interval [0,2] where f '(c) = average rate of change. In other words, there exists a point c in the interval [0,2] such that f '(c) = 4.

To determine if any critical points exist, we need to solve f '(x) = 4. Let's set the derivative equal to 4 and solve for x:

3(x-2)^2 = 4
Divide both sides by 3:
(x-2)^2 = 4/3

Taking the square root of both sides:
x-2 = ±√(4/3)

Adding 2 to both sides:
x = 2 ± √(4/3)

Therefore, there are two values of c that satisfy f '(c) = 4, which are c = 2 + √(4/3) and c = 2 - √(4/3). The mean value theorem predicts that at these values of c, f(x) has an instantaneous rate of change equal to the average rate of change over the interval [0,2].

I hope this helps you understand how to approach these types of questions. Let me know if you have any more questions!