Mean Value Theorem Calculator
Apply the Mean Value Theorem to find points on a curve
Supported: +, -, *, /, **, sin, cos, tan, sqrt, log, ln, exp, abs, pi, e
Mean Value Theorem Formula:
f'(c) = rac{f(b) - f(a)}{b - a}where a < c < b
Enter function and interval, then click Calculateto see result.
Enter function and interval, then click Calculateto see result.
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How to Use This Calculator
Step-by-step guide to get accurate results
What is a Mean Value Theorem Calculator?
How to Use the Calculator
Enter the Function
Enter the Interval [a, b]
Click Calculate
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Key Features of the Calculator
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Instant Calculations
Supports Multiple Functions
Easy-to-Use Interface
Accurate Results
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Example Calculations
Example 1
Example 2
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The Mean Value Theorem (MVT) is a key result in differential calculus. It states that if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one point c in (a, b) such that the derivative f'(c) equals the average rate of change over [a, b].
Geometrically, this means there is at least one point where the tangent to the curve is parallel to the secant line joining (a, f(a)) and (b, f(b)).
Mean Value Theorem Formula:
f'(c) = rac{f(b) - f(a)}{b - a}where a < c < b
Step-by-Step Logic:
- Verify the function is continuous on [a, b] and differentiable on (a, b).
- Compute the average rate of change: rac{f(b) - f(a)}{b - a}.
- Find the derivative f'(x).
- Solve f'(c) = rac{f(b) - f(a)}{b - a} for c in (a, b).
Conditions for MVT:
- f(x) must be continuous on the closed interval [a,b]
- f(x) must be differentiable on the open interval (a,b)
- The theorem guarantees at least one such c, but there may be more.
- Enter your function f(x) using standard mathematical notation (e.g., x**2, sin(x), -4*x**3 + 6*x - 2).
- Specify the interval endpoints a and b (with a < b).
- Click "Calculate Point" to find the value(s) of c.
- Review the step-by-step solution showing all calculations.
- Verify that each c lies in the open interval (a, b).
Example 1: f(x) = x², [1, 3]
Step-by-step solution:
• f(1) = 1, f(3) = 9
• AROC = (9-1)/(3-1) = 4
• f'(x) = 2x
• Solve: 2c = 4 ⇒ c = 2
• Verify: 2 ∈ (1,3) ✅
Example 2: f(x) = x³, [0, 2]
Step-by-step solution:
• f(0) = 0, f(2) = 8
• AROC = (8-0)/(2-0) = 4
• f'(x) = 3x²
• Solve: 3c² = 4 ⇒ c² = 4/3 ⇒ c = ±√(4/3) ≈ ±1.154 (only positive in interval)
• Verify: 1.154 ∈ (0,2) ✅
Example 3: f(x) = -4x³ + 6x - 2, [-4, 2] (Corrected Calculation)
Step-by-step solution:
• f(-4) = 230, f(2) = -22
• AROC = (-22 - 230)/(2 - (-4)) = -252/6 = -42
• f'(x) = -12x² + 6
• Solve: -12c² + 6 = -42 ⇒ -12c² = -48 ⇒ c² = 4 ⇒ c = ±2
• Verify: -2 and 2 ∈ (-4,2)? -2 ✅, 2 ❌ (2 is endpoint, but open interval)