Math Problem Statement

In the following exercises, use the Intermediate Value

Theorem (IVT). Let h(x) =

3x2 − 4, x ≤ 2

5 + 4x, x > 2

Over the interval [0, 4], there is no value of x such that h(x) = 10,

although h(0) < 10 and h(4) > 10. Explain why this does not contradict the IVT.

Solution

The Intermediate Value Theorem (IVT) states that if a function f(x)f(x) is continuous on a closed interval [a,b][a, b] and f(a)f(a) and f(b)f(b) take on opposite signs (i.e., f(a)<k<f(b)f(a) < k < f(b) or f(b)<k<f(a)f(b) < k < f(a) for some value kk), then there exists at least one value cc in the interval (a,b)(a, b) such that f(c)=kf(c) = k.

To analyze whether the IVT applies here, let's break down the given function h(x)h(x):

\begin{cases} 3x^2 - 4 & \text{for } x \leq 2 \\ 5 + 4x & \text{for } x > 2 \end{cases}$$ ### Step 1: Check continuity of $$h(x)$$ over the interval [0, 4] To apply the IVT, $$h(x)$$ must be continuous on the entire interval [0, 4]. However, the function is defined piecewise, with one expression for $$x \leq 2$$ and a different expression for $$x > 2$$. We need to check if the function is continuous at the point where the pieces meet, which is at $$x = 2$$. For continuity at $$x = 2$$, the left-hand limit (as $$x \to 2^-$$) and the right-hand limit (as $$x \to 2^+$$) must be equal, and the function's value at $$x = 2$$ must match those limits. - For $$x \leq 2$$, $$h(x) = 3x^2 - 4$$. So, $$h(2) = 3(2)^2 - 4 = 12 - 4 = 8$$. - For $$x > 2$$, $$h(x) = 5 + 4x$$. The right-hand limit as $$x \to 2^+$$ is $$h(2^+) = 5 + 4(2) = 5 + 8 = 13$$. Clearly, $$h(2^-) = 8$$ and $$h(2^+) = 13$$, which means that the function has a jump discontinuity at $$x = 2$$. ### Step 2: Why the IVT does not apply The IVT requires the function to be continuous over the entire interval [0, 4], but $$h(x)$$ is not continuous at $$x = 2$$. Therefore, the IVT does not apply to the interval [0, 4], and we cannot conclude that there is a value $$c$$ such that $$h(c) = 10$$ using the IVT. ### Conclusion Although $$h(0) < 10$$ and $$h(4) > 10$$, there is no contradiction with the Intermediate Value Theorem because the function is not continuous on the entire interval [0, 4] due to the jump discontinuity at $$x = 2$$. For the IVT to hold, the function must be continuous throughout the interval, which is not the case here. --- Would you like more details or have any questions? Here are 5 questions to further explore the concept: 1. What are the conditions necessary for applying the Intermediate Value Theorem? 2. How can you tell if a piecewise function is continuous at a particular point? 3. What is a jump discontinuity, and how does it affect the application of the IVT? 4. Can you think of a scenario where the IVT would apply successfully for this function on a different interval? 5. How could you modify the function $$h(x)$$ to make it continuous at $$x = 2$$? **Tip:** Always check for continuity when applying the IVT to ensure the theorem’s conditions are satisfied.

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Math Problem Analysis

Mathematical Concepts

Calculus
Continuity
Piecewise Functions

Formulas

h(x) = 3x^2 − 4 for x ≤ 2
h(x) = 5 + 4x for x > 2

Theorems

Intermediate Value Theorem

Suitable Grade Level

Grades 11-12, College Introductory Calculus

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