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Let f be a function with first derivative defined by f'(x)= 3x^2-6/x^2 for x>0. It is known that f(1)=9 and f(3)= 11. What value of x in the open interval (1,3) satisfies the conclusion of the Mean Interval Theorem for f on the closed interval [1,3]?

A. sq root 6
B. sq root 3
C. sq root 2
D. 1

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General Formulas and Concepts

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Mean Value Theorem:
[tex]\displaystyle f'(c) = \frac{f(b) - f(a)}{b - a}[/tex]

  1. [Condition] f(x) is continuous on the closed interval [a, b]
  2. [Condition] f(x) is differentiable on the open interval [a, b]

Application

Step 1: Define

Let's organize what the problem gives us:

We are given a function with the first derivative with a restriction:
[tex]\displaystyle f'(x) = \frac{3x^2 - 6}{x^2} ,\ x > 0[/tex]

We are also given several points from the function itself:

  • f(1) = 9
  • f(3) = 11

Lastly, we are given the interval [1, 3].

Step 2: Verify

We must first verify if we can use the Mean Value Theorem. Recall the conditions needed to be met in order to use the theorem.

For the 1st condition, we are told from the problem that f(x) is continuous on the closed interval [1, 3] because the first derivative (and subsequently the function) is restrained for x > 0.

∴ f(x) is continuous on the closed interval [1, 3].

For the 2nd condition, we are told from the problem that f(x) is differentiable on the open interval (1, 3).

∴ f(x) is differentiable on the open interval (1, 3).

Step 3: Solving

Now that we know that the conditions for Mean Value Theorem are met, we can apply it to solve our question.

Let's first use our Mean Value Theorem Formula:

[tex]\displaystyle\begin{aligned}f'(c) & = \frac{f(3) - f(1)}{3 - 1} \\& = \frac{11 - 9}{2} \\& = \frac{2}{2} \\& = \boxed{ 1 } \end{aligned}[/tex]

∴ the Mean Value Theorem tells us that the derivative at some point c (in the problem's case, x) must equal the value of 1.

Now we substitute f'(c) = 1 into our defined first derivative and solve:

[tex]\displaystyle\begin{aligned}1 & = \frac{3x^2 - 6}{x^2} \\x^2 & = 3x^2 - 6 \\-2x^2 & = -6 \\x^2 & = 3 \\x & = \boxed{ \pm \sqrt{3} } \\\end{aligned}[/tex]

Since our first derivative is restrained by x > 0, we ultimately end up with one root:

[tex]\displaystyle x = \pm \sqrt{3} \longrightarrow \boxed{ x = \sqrt{3} }[/tex]

∴ [tex]\displaystyle \boxed{ c = \sqrt{3} }[/tex]

Answer

∴ the value of x in the open interval (1, 3) that satisfies the conclusion of the Mean Value Theorem for f on the closed interval [1, 3] is equal to B. √3.

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Learn more about derivatives: https://brainly.com/question/30070254

Learn more about Calculus: https://brainly.com/question/30043419

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Topic: Calculus I

Unit: Differentiation

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To find the value of x that satisfies the conclusion of the Mean Value Theorem for f on the closed interval [1,3], we need to find a value of x in the open interval (1,3) such that f'(x) equals the average rate of change of f on the interval [1,3].

The average rate of change of f on the interval [1,3] is equal to (f(3) - f(1)) / (3 - 1) = (11 - 9) / 2 = 1.

Substituting this value into the equation f'(x) = 3x^2 - 6 / x^2, we get 1 = 3x^2 - 6 / x^2. Solving for x, we find that x = sq root 3. Therefore, the answer is B. sq root 3.

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