Math Problem Statement

determine if mean value theorem can be applied to the function f(x)=x^2 on the closed interval [3,9]. if it can be applied, find all numbers c in the open interval (3,9) such that f'(c)=(f(9)-f(3))/(9-3)

Solution

Let's analyze the problem step by step.

Step 1: Check if the Mean Value Theorem (MVT) can be applied to f(x)=x2f(x) = x^2 on the interval [3,9][3, 9].

The Mean Value Theorem states that if f(x)f(x) is continuous on a closed interval [a,b][a, b] and differentiable on the open interval (a,b)(a, b), then there exists at least one point cc in (a,b)(a, b) such that:

f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}

For f(x)=x2f(x) = x^2:

  • Continuity: f(x)=x2f(x) = x^2 is a polynomial function, and polynomials are continuous for all real values of xx. Therefore, f(x)f(x) is continuous on the closed interval [3,9][3, 9].
  • Differentiability: The derivative of f(x)=x2f(x) = x^2 is f(x)=2xf'(x) = 2x, which is also continuous and differentiable for all real xx. Therefore, f(x)f(x) is differentiable on the open interval (3,9)(3, 9).

Since the function meets both the continuity and differentiability conditions, the Mean Value Theorem can be applied.

Step 2: Find the value of f(9)f(3)93\frac{f(9) - f(3)}{9 - 3}.

We need to calculate the average rate of change of f(x)f(x) on the interval [3,9][3, 9]:

f(9)=92=81f(9) = 9^2 = 81 f(3)=32=9f(3) = 3^2 = 9 f(9)f(3)93=81993=726=12\frac{f(9) - f(3)}{9 - 3} = \frac{81 - 9}{9 - 3} = \frac{72}{6} = 12

So, we want to find c(3,9)c \in (3, 9) such that f(c)=12f'(c) = 12.

Step 3: Solve for cc where f(c)=12f'(c) = 12.

We know that f(x)=2xf'(x) = 2x. Therefore, we need to solve:

2c=122c = 12 c=6c = 6

Conclusion:

The number c=6c = 6 satisfies the Mean Value Theorem for the function f(x)=x2f(x) = x^2 on the interval [3,9][3, 9].

Summary:

  • The Mean Value Theorem can be applied to f(x)=x2f(x) = x^2 on the interval [3,9][3, 9] because the function is continuous and differentiable on the appropriate intervals.
  • The value of cc that satisfies f(c)=f(9)f(3)93f'(c) = \frac{f(9) - f(3)}{9 - 3} is c=6c = 6.

Do you need any further clarification or additional details? Here are a few related questions:

  1. How does the Mean Value Theorem apply to other types of functions?
  2. What would happen if f(x)f(x) were not continuous on the interval?
  3. Can the Mean Value Theorem be applied to non-differentiable functions?
  4. How can you interpret the geometric meaning of the Mean Value Theorem?
  5. What are some examples of functions where the MVT does not apply?

Tip: When using the Mean Value Theorem, always check first if the function is continuous on the closed interval and differentiable on the open interval before attempting to find the value of cc.

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Math Problem Analysis

Mathematical Concepts

Calculus
Continuity
Differentiability
Mean Value Theorem

Formulas

f'(c) = (f(b) - f(a)) / (b - a)
f'(x) = 2x (derivative of f(x) = x^2)

Theorems

Mean Value Theorem

Suitable Grade Level

Grades 11-12