Math Problem Statement

Determine whether the Intermediate Value Theorem guarantees that the function k(x) = 2x^3 - 11x^2 + 6x + 6 has a zero on the given interval.

Solution

The problem asks you to determine whether the Intermediate Value Theorem (IVT) guarantees that the function k(x)=2x311x2+6x+6k(x) = 2x^3 - 11x^2 + 6x + 6 has a zero on the given interval, using different intervals like [1, 2], [2, 3], etc.

The Intermediate Value Theorem (IVT)

The IVT states that if a function f(x)f(x) is continuous on a closed interval [a, b], and f(a)f(a) and f(b)f(b) have opposite signs (i.e., one is positive and the other is negative), then there must be at least one cc in the interval such that f(c)=0f(c) = 0. In other words, the function must cross the x-axis at some point within the interval.

Step-by-step explanation:

  1. The problem specifically asks you to find values of k(1)k(1) and k(2)k(2) to see if they have opposite signs.

  2. We are given k(x)=2x311x2+6x+6k(x) = 2x^3 - 11x^2 + 6x + 6, so we can compute: k(1)=2(1)311(1)2+6(1)+6=211+6+6=3k(1) = 2(1)^3 - 11(1)^2 + 6(1) + 6 = 2 - 11 + 6 + 6 = 3 k(2)=2(2)311(2)2+6(2)+6=2(8)11(4)+12+6=1644+12+6=10k(2) = 2(2)^3 - 11(2)^2 + 6(2) + 6 = 2(8) - 11(4) + 12 + 6 = 16 - 44 + 12 + 6 = -10

  3. Since k(1)=3k(1) = 3 (positive) and k(2)=10k(2) = -10 (negative), they have opposite signs.

  4. By the Intermediate Value Theorem, because the function k(x)k(x) is continuous on the interval [1, 2] and k(1)k(1) and k(2)k(2) have opposite signs, there is at least one zero of the function within this interval.

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Questions to think about:

  1. What are the exact conditions for the Intermediate Value Theorem to hold?
  2. How can you check continuity for a given function?
  3. What would happen if k(1)k(1) and k(2)k(2) had the same sign?
  4. How would you approach other intervals like [2, 3] or [3, 4] in this problem?
  5. Can the Intermediate Value Theorem tell you where exactly the zero is?

Tip: Always make sure the function is continuous before applying the IVT!

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Math Problem Analysis

Mathematical Concepts

Intermediate Value Theorem
Polynomials
Continuity
Roots of Functions

Formulas

k(x) = 2x^3 - 11x^2 + 6x + 6

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 10-12