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Free Intermediate Value Theorem calculator for checking theorem conditions, finding roots, and verifying value existence with step-by-step solutionsfor calculus and mathematical analysis. Perfect for students learning continuity and root-finding.
Last updated: February 2, 2026
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Theorem Applies:
Yes
Intermediate Value Theorem Check
Formula: f(a) < k < f(b) or f(b) < k < f(a)
f(a):
-2.000
f(b):
2.000
Root Exists:
Yes
Condition
f(a) < k < f(b) or f(b) < k < f(a)
Check if theorem applies to guarantee solution existence
Method
Bisection method approximation
Use IVT to find roots when f(a) and f(b) have opposite signs
Condition
k between f(a) and f(b)
Verify if a specific value can be achieved in the interval
Example
Temperature reaches every value between two points
Proving temperature continuity in physical systems
Example
Population reaches intermediate values
Modeling population growth and decay processes
Example
Supply and demand curves intersect
Proving market equilibrium exists
Intermediate Value Theorem with f(x) = x² - 2, [0, 2], k = 0:
Theorem Applies
Yes
f(0)
-2.000
f(2)
2.000
The Intermediate Value Theorem is a fundamental principle in calculus that guarantees the existence of solutions under certain conditions. Understanding this theorem is essential for calculus, mathematical analysis, and numerical methodswhere continuity and solution existence are important.
This systematic approach ensures accurate Intermediate Value Theorem calculations for any continuous function.
The theorem requires two main conditions: 1) The function f must be continuous on the closed interval [a, b], and 2) The target value k must be between f(a) and f(b). If both conditions are met, then there exists at least one number c in the open interval (a, b) such that f(c) = k. The theorem guarantees existence but not uniqueness of solutions.
Need help with other calculus topics? Check out our derivative calculator and concavity calculator.
Get Custom Calculator for Your PlatformFunction: f(x) = x² - 2
Interval: [0, 2]
Target value: k = 0
Calculation type: Theorem check
Final Results:
Theorem Applies
Yes
Root Exists
Yes
f(0)
-2.000
f(2)
2.000
f(x) = x³ - 3x + 1, [0, 2], k = 0
Root exists: c ≈ 1.532
f(x) = sin(x), [0, π], k = 0.5
Value exists: c ≈ 0.524
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