Mean Value Theorem and Rolle's Theorem Application

MathematicsCalculusHardSTEM

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17. Let $f$ be a function that is continuous on $[1, 2]$ and differentiable on $(1, 2)$ with $f(1) = 2$ and $f(2) = 4$, and $g(x) = rac{f(x)}{x}$. Which of the following statements are true?

I. There exists at least one $c \in (1, 2)$ such that $f'(c) = 2$.

II. There exists at least one $c \in (1, 2)$ such that $g'(c) = 0$.

III. There exists at least one $c \in (1, 2)$ such that $f'(c) = rac{f(c)}{c}$.

(a) I only

(b) II only

(c) I, II

(d) I, III

(e) I, II, III

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Step by Step Written Solution

1
Step 1

Hi Mustafa, let's solve this calculus problem involving the Mean Value Theorem. We are given a function f that is continuous and differentiable on the interval from 1 to 2, and a helper function g of x.

Given Information

$$f \text{ is continuous on } [1, 2]$$
$$f \text{ is differentiable on } (1, 2)$$
$$f(1) = 2, \quad f(2) = 4$$
$$g(x) = \frac{f(x)}{x}$$
2
Step 2

Let's investigate statement one first. It asks if there exists a point c where the derivative f prime of c equals 2.

Analyzing Statement I

$$I. \exists c \in (1, 2) \text{ such that } f'(c) = 2$$
3
Step 3

Since f is continuous and differentiable, we can apply the Mean Value Theorem. The theorem states there is a c in the interval such that the derivative equals the average rate of change.

$$f'(c) = \frac{f(2) - f(1)}{2 - 1}$$
4
Step 4

Plugging in the given values, we get four minus two over two minus one.

5
Step 5

This simplifies to two over one, which is two. So, statement one is definitely true.

6
Step 6

Now let's look at statement two. It asks if there is a c such that the derivative of g of x equals zero.

Analyzing Statement II

$$II. \exists c \in (1, 2) \text{ such that } g'(c) = 0$$
$$g(x) = \frac{f(x)}{x}$$
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Step 7

We check the values of g at the endpoints. First, g of one equals f of one divided by one, which is two.

$$g(1) = \frac{f(1)}{1} = \frac{2}{1} = 2$$
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Step 8

Next, g of two equals f of two divided by two. Substituting the value four for f of two, we get four divided by two, which is also two.

$$g(2) = \frac{f(2)}{2} = \frac{4}{2} = 2$$
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Step 9

Since g of one equals g of two, and g is continuous and differentiable on the interval, we can apply Rolle's Theorem.

Since $g(1) = g(2) = 2$, by Rolle's Theorem:

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Step 10

Rolle's Theorem guarantees that there exists at least one c in the open interval such that the derivative of g equals zero. Thus, statement two is true as well.

$$g'(c) = 0 \quad \text{for some } c \in (1, 2)$$

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About This Question

Subject
Mathematics
Topic
Calculus
Difficulty
Hard
Exam
STEM
Question Type
Multiple Choice

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