Math Problem Statement

Consider the function f(x) which is continuous on the interval [0,6] and where f(0) = 12 and f(6) = 7. Based on this information, will there be a value x = a on the interval [0,6] such that f(a) = 6?

Solution

Here’s a summary of the questions and their answers from the provided image:


Top Row

  1. Question: Given f(x)f(x) continuous on [0,6][0,6], f(0)=12f(0)=12, f(6)=7f(6)=7, will f(a)=6f(a)=6 exist in [0,6][0,6]?

    • Answer: Yes, by the Intermediate Value Theorem (IVT), since f(x)f(x) is continuous and f(0)=12>6>f(6)=7f(0)=12 > 6 > f(6)=7.
    • Answer Key: 0.327
  2. Question: For h(x)=3xπ+cosxh(x) = \frac{3x}{\pi} + \cos x, apply the Mean Value Theorem (MVT) on [0,π2][0, \frac{\pi}{2}].

    • Answer: x=cx = c exists in (0,π2)(0, \frac{\pi}{2}), and it satisfies h(c)=h(π/2)h(0)π/2h'(c) = \frac{h(\pi/2)-h(0)}{\pi/2}.
    • Answer Key: 4

Middle Row

  1. Question: Given table values for f(x)f(x), where x=1,3,4,7,11x=1, 3, 4, 7, 11, find the fewest number of times f(x)=50f(x) = 50.

    • Answer: Using the IVT, check intervals between f(x)f(x) values.
    • Answer Key: 0.327
  2. Question: Given f(x)=x44x3+8x2+6x20f(x) = x^4 - 4x^3 + 8x^2 + 6x - 20, table completion and zero existence check.

    • Answer: The table will determine whether f(x)f(x) changes sign (indicating a zero).
    • Answer Key: 0.226

Bottom Row

  1. Question: For f(x)=x+2f(x) = \sqrt{x+2} on [3,0][-3,0], determine if MVT conditions are met.

    • Answer: Yes, f(x)f(x) is continuous and differentiable. The guaranteed cc value satisfies MVT.
    • Answer Key: 0.690
  2. Question: Find the minimum value for p(x)=x224xp(x) = -x^2 - 24x on [6,1][-6,-1].

    • Answer: Use endpoints and derivative test: p(x)p(x) reaches its minimum at x=4x=-4.
    • Answer Key: -4

Let me know if you’d like detailed solutions or explanations for any of these problems!

5 Related Questions:

  1. How does the Intermediate Value Theorem guarantee a solution?
  2. What is the Mean Value Theorem, and how is it applied?
  3. How can you determine zeros of a function from a table of values?
  4. How do you use derivatives to find minimum or maximum values of a function?
  5. What are the conditions for applying the Mean Value Theorem?

Tip:

For verifying f(x)=kf(x)=k in an interval, use Intermediate Value Theorem: check for continuity and ensure f(a)kf(b)f(a) \leq k \leq f(b).

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Math Problem Analysis

Mathematical Concepts

Continuity
Intermediate Value Theorem

Formulas

Intermediate Value Theorem: If a function f(x) is continuous on [a,b] and f(a) < N < f(b), then there exists at least one c ∈ (a, b) such that f(c) = N.

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 11-12