Math Problem Statement
Use the Intermediate Value Theorem to show that there is a solution to the given equation in the specified interval: i. [4 points] ln(x) = 4 − √ x, for x ∈ (4,9). 1 ii. [4 points] e x = 3 − 2x, for x ∈ (0,1)
Solution
The Intermediate Value Theorem (IVT) states that if a function is continuous on a closed interval , and if and have opposite signs, then there exists at least one such that .
Part i:
We can rewrite the equation as a function:
We aim to find if there exists such that , by applying the IVT.
Step 1: Check Continuity
- Both and are continuous for .
- Therefore, is also continuous on .
Step 2: Evaluate at the endpoints of the interval
-
At : Using , we get:
-
At : Using , we get:
Step 3: Apply the IVT
- and , so and .
- Since is continuous and changes sign between and , the IVT guarantees that there is at least one solution such that .
Conclusion for Part i:
There exists a solution to the equation in the interval .
Part ii:
We rewrite the equation as a function:
We aim to show there is a solution such that , using the IVT.
Step 1: Check Continuity
- Both and are continuous for all .
- Therefore, is continuous on .
Step 2: Evaluate at the endpoints of the interval
-
At :
-
At : Using , we get:
Step 3: Apply the IVT
- and , so and .
- Since is continuous and changes sign between and , the IVT guarantees that there is at least one solution such that .
Conclusion for Part ii:
There exists a solution to the equation in the interval .
Do you want more details on any step, or further explanation?
Here are some related questions to deepen your understanding:
- How can the Intermediate Value Theorem be applied to functions with discontinuities?
- What are some common cases where the Intermediate Value Theorem does not apply?
- How can we estimate the value of the solution in each case using numerical methods?
- Can the Intermediate Value Theorem guarantee more than one solution in an interval?
- How does the continuity of a function affect the applicability of the IVT?
Tip: The IVT is useful not only for proving the existence of a solution but also for narrowing down intervals where a root might lie, which can then be solved more precisely using other methods like Newton's method or the bisection method.
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Math Problem Analysis
Mathematical Concepts
Calculus
Continuity
Intermediate Value Theorem
Formulas
f(x) = ln(x) - (4 - √x)
g(x) = e^x - (3 - 2x)
Theorems
Intermediate Value Theorem
Suitable Grade Level
Grades 11-12, College
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