Math Problem Statement

The figure shows the graph of the function f on the interval [2,8] together with the dashed line y = -2. We see that f(2) < -2 and f(8) > -2 but f(x) ≠ -2 anywhere on [2,8]. Why does this NOT violate the Intermediate Value Theorem?

Solution

The problem asks why the given function f(x)f(x) on the interval [2,8][2, 8], with the condition f(2)<2f(2) < -2, f(8)>2f(8) > -2, but f(x)2f(x) \neq -2 anywhere in the interval, does not violate the Intermediate Value Theorem (IVT).

Key points:

  1. Intermediate Value Theorem states that if a function f(x)f(x) is continuous on a closed interval [a,b][a, b], and if LL is any value between f(a)f(a) and f(b)f(b), then there exists some c(a,b)c \in (a, b) such that f(c)=Lf(c) = L.

  2. Here, the function is continuous on the interval [2,8][2, 8] and we have:

    • f(2)<2f(2) < -2
    • f(8)>2f(8) > -2
    • f(x)2f(x) \neq -2 for any x[2,8]x \in [2, 8]

Why it doesn't violate IVT:

  • The IVT requires that there is some value cc where f(c)=Lf(c) = L if LL lies between f(a)f(a) and f(b)f(b).
  • However, the value L=2L = -2 does not occur anywhere in the function's range, even though the function crosses from below 2-2 to above 2-2 as shown by the graph. This can happen because the function could "jump" over the value 2-2 or there is a vertical asymptote or discontinuity at x=2x = -2, despite being continuous at all other points.

This situation does not violate IVT because the theorem only guarantees that a value is taken if the function is continuous and takes all intermediate values between f(a)f(a) and f(b)f(b). Here, the critical value 2-2 is avoided even though the function’s outputs exceed and go below 2-2.


Would you like more details on the Intermediate Value Theorem or the nature of continuous functions?

Here are 5 questions to expand on the topic:

  1. What are the conditions for the Intermediate Value Theorem to hold?
  2. Can a function be continuous but not achieve certain values within an interval?
  3. What happens if a function has a discontinuity within the interval of IVT?
  4. How can a function "skip" certain values despite being continuous?
  5. How is continuity related to real-world problems like this graph?

Tip: Understanding how IVT applies requires knowing how continuity ensures the function covers all values between the endpoints but does not guarantee it hits all specific values in between!

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Math Problem Analysis

Mathematical Concepts

Calculus
Intermediate Value Theorem
Continuity

Formulas

-

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grade 12 - Early College