Mean Value Theorem

Understanding the theoretical foundation connecting average and instantaneous rates

🎓 Mean Value Theorem

What is the Mean Value Theorem?

The Mean Value Theorem (MVT) is one of the most important theoretical results in calculus. It connects the average rate of change over an interval to the instantaneous rate of change at some point.

💡 Key Idea: If you drive 100 miles in 2 hours (average 50 mph), at some moment you were going exactly 50 mph!


The Theorem (Formal Statement)

Let ff be a function that is:

  1. Continuous on the closed interval [a,b][a, b]
  2. Differentiable on the open interval (a,b)(a, b)

Then there exists at least one number cc in (a,b)(a, b) such that:

f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}

In Words

There is at least one point where the instantaneous rate of change (derivative) equals the average rate of change (slope of secant line).


Visual Understanding

The Secant Line

The secant line connects (a,f(a))(a, f(a)) to (b,f(b))(b, f(b)).

Its slope is: f(b)f(a)ba\frac{f(b) - f(a)}{b - a}

The Tangent Line

The MVT says there's a point cc where the tangent line is parallel to the secant line.

Geometric interpretation: Somewhere between aa and bb, the curve has a tangent line with the same slope as the overall average slope.


Rolle's Theorem (Special Case)

Rolle's Theorem is a special case of MVT when f(a)=f(b)f(a) = f(b).

Statement

If ff is continuous on [a,b][a, b], differentiable on (a,b)(a, b), and f(a)=f(b)f(a) = f(b), then there exists cc in (a,b)(a, b) such that:

f(c)=0f'(c) = 0

In Words

If a function starts and ends at the same height, somewhere in between it has a horizontal tangent!

Example: If you hike up and back down to your starting elevation, at some point you reached the top (where slope = 0).


How to Apply the MVT

Step-by-Step Process

Step 1: Verify the hypotheses

  • Is ff continuous on [a,b][a, b]?
  • Is ff differentiable on (a,b)(a, b)?

Step 2: Calculate the average rate of change m=f(b)f(a)bam = \frac{f(b) - f(a)}{b - a}

Step 3: Find f(x)f'(x)

Step 4: Solve f(c)=mf'(c) = m for cc

Step 5: Verify that cc is in (a,b)(a, b)


Example 1: Finding the Point

Find all values of cc that satisfy the MVT for f(x)=x2f(x) = x^2 on [1,3][1, 3].

Step 1: Check hypotheses

f(x)=x2f(x) = x^2 is a polynomial:

  • Continuous everywhere ✓
  • Differentiable everywhere ✓

MVT applies!


Step 2: Average rate of change

f(3)f(1)31=912=82=4\frac{f(3) - f(1)}{3 - 1} = \frac{9 - 1}{2} = \frac{8}{2} = 4


Step 3: Find derivative

f(x)=2xf'(x) = 2x


Step 4: Solve f(c)=4f'(c) = 4

2c=42c = 4 c=2c = 2


Step 5: Verify

Is c=2c = 2 in (1,3)(1, 3)? Yes! ✓

Answer: c=2c = 2 satisfies the MVT.

At x=2x = 2, the instantaneous rate equals the average rate!


Example 2: MVT Doesn't Apply

Why doesn't MVT apply to f(x)=xf(x) = |x| on [1,1][-1, 1]?

ff is continuous on [1,1][-1, 1]

But ff is NOT differentiable at x=0x = 0 (corner/cusp) ✗

Since 00 is in (1,1)(-1, 1), the MVT does not apply.


Why the Hypotheses Matter

Continuity Required

If ff has a jump discontinuity, there may be no point where the tangent slope equals the average slope.

Differentiability Required

If ff has a corner (like x|x| at x=0x=0), the derivative doesn't exist there, so we can't apply MVT.

On (a,b)(a, b), not [a,b][a, b]

The function only needs to be differentiable on the open interval. It's okay if f(a)f'(a) or f(b)f'(b) don't exist!


Applications of MVT

Application 1: Proving Functions are Equal

If f(x)=g(x)f'(x) = g'(x) for all xx in an interval, then: f(x)=g(x)+Cf(x) = g(x) + C

for some constant CC.

Why: If h(x)=f(x)g(x)h(x) = f(x) - g(x), then h(x)=0h'(x) = 0 everywhere. By MVT, hh must be constant!

Application 2: Estimating Values

If we know f(x)f'(x) is bounded, we can estimate how much ff can change.

Example: If f(x)2|f'(x)| \leq 2 for all xx in [0,5][0, 5], then: f(5)f(0)25=10|f(5) - f(0)| \leq 2 \cdot 5 = 10

Application 3: Speed Limits

If a car travels 120 miles in 2 hours (average 60 mph), by MVT the car must have been going exactly 60 mph at some instant!

If the speed limit is 55 mph, the driver was definitely speeding at some point. 🚗


Increasing and Decreasing Functions

Important Corollary

Let ff be continuous on [a,b][a, b] and differentiable on (a,b)(a, b).

Increasing Function Test:

  • If f(x)>0f'(x) > 0 for all xx in (a,b)(a, b), then ff is increasing on [a,b][a, b]

Decreasing Function Test:

  • If f(x)<0f'(x) < 0 for all xx in (a,b)(a, b), then ff is decreasing on [a,b][a, b]

Constant Function Test:

  • If f(x)=0f'(x) = 0 for all xx in (a,b)(a, b), then ff is constant on [a,b][a, b]

Proof idea: Uses MVT repeatedly!


⚠️ Common Mistakes

Mistake 1: Wrong Interval

The value cc must be in the open interval (a,b)(a, b), not including endpoints!

Mistake 2: Not Checking Hypotheses

Always verify continuity and differentiability before applying MVT!

Mistake 3: Expecting Unique cc

MVT says "at least one" cc exists. There might be multiple values!

Mistake 4: Confusing with Extreme Value Theorem

  • EVT: Guarantees max and min exist
  • MVT: Guarantees a point where instantaneous rate = average rate

MVT vs. Other Theorems

Extreme Value Theorem (EVT)

EVT: Continuous function on [a,b][a, b] has absolute max and min

MVT: Connects average and instantaneous rates

Intermediate Value Theorem (IVT)

IVT: Continuous function takes all values between f(a)f(a) and f(b)f(b)

MVT: About derivatives, not function values


Extended Mean Value Theorem (Cauchy)

There's a more general version for two functions:

If ff and gg are continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then there exists cc in (a,b)(a, b) such that:

f(c)g(c)=f(b)f(a)g(b)g(a)\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}

This is used in proving L'Hôpital's Rule!


Proof Sketch of MVT

Idea: Create a function that measures the vertical distance from the curve to the secant line, then apply Rolle's Theorem.

Define: h(x)=f(x)[f(a)+f(b)f(a)ba(xa)]h(x) = f(x) - \left[f(a) + \frac{f(b)-f(a)}{b-a}(x-a)\right]

Then h(a)=h(b)=0h(a) = h(b) = 0, so by Rolle's Theorem, there exists cc where h(c)=0h'(c) = 0.

Working out h(c)=0h'(c) = 0 gives f(c)=f(b)f(a)baf'(c) = \frac{f(b)-f(a)}{b-a}


📝 Key Takeaways

  1. MVT connects average and instantaneous rates - somewhere the tangent is parallel to the secant

  2. Requires continuity on [a,b][a, b] and differentiability on (a,b)(a, b)

  3. Guarantees existence of at least one cc, but doesn't tell you how many

  4. Applications: Proving functions equal, bounding changes, theoretical foundation

  5. Geometric meaning: Tangent line parallel to secant line

  6. Rolle's Theorem is the special case when f(a)=f(b)f(a) = f(b)


Practice Tips

  1. Always check hypotheses before applying MVT
  2. Calculate average rate first: f(b)f(a)ba\frac{f(b)-f(a)}{b-a}
  3. Set f(c)f'(c) equal to the average rate and solve
  4. Verify cc is in the open interval (a,b)(a, b)
  5. Understand geometrically - it's about parallel slopes!

📚 Practice Problems

1Problem 1medium

Question:

Verify that f(x)=x33x+1f(x) = x^3 - 3x + 1 satisfies the hypotheses of the Mean Value Theorem on [0,2][0, 2], then find all values of cc that satisfy the conclusion.

💡 Show Solution

Step 1: Verify hypotheses

f(x)=x33x+1f(x) = x^3 - 3x + 1 is a polynomial.

Polynomials are continuous everywhere ✓

Polynomials are differentiable everywhere ✓

MVT applies on [0,2][0, 2]


Step 2: Calculate average rate of change

f(0)=033(0)+1=1f(0) = 0^3 - 3(0) + 1 = 1

f(2)=233(2)+1=86+1=3f(2) = 2^3 - 3(2) + 1 = 8 - 6 + 1 = 3

Average rate: f(2)f(0)20=312=22=1\frac{f(2) - f(0)}{2 - 0} = \frac{3 - 1}{2} = \frac{2}{2} = 1


Step 3: Find f(x)f'(x)

f(x)=3x23f'(x) = 3x^2 - 3


Step 4: Solve f(c)=1f'(c) = 1

3c23=13c^2 - 3 = 1

3c2=43c^2 = 4

c2=43c^2 = \frac{4}{3}

c=±23=±233c = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}


Step 5: Check which values are in (0,2)(0, 2)

c=2331.155c = \frac{2\sqrt{3}}{3} \approx 1.155 ✓ (in (0,2)(0, 2))

c=2331.155c = -\frac{2\sqrt{3}}{3} \approx -1.155 ✗ (not in (0,2)(0, 2))

Answer: c=233c = \frac{2\sqrt{3}}{3} satisfies the Mean Value Theorem.

2Problem 2easy

Question:

Explain why the Mean Value Theorem does not apply to f(x)=1xf(x) = \frac{1}{x} on [1,1][-1, 1].

💡 Show Solution

Check Hypothesis 1: Continuity on [1,1][-1, 1]

f(x)=1xf(x) = \frac{1}{x} has a vertical asymptote at x=0x = 0.

ff is NOT continuous at x=0x = 0

Since 0[1,1]0 \in [-1, 1], the function is not continuous on the entire interval.


Conclusion

The Mean Value Theorem does not apply because ff is not continuous on [1,1][-1, 1].


Additional observation:

Even if we tried to apply it:

f(1)=1f(-1) = -1 and f(1)=1f(1) = 1

Average rate: 1(1)1(1)=22=1\frac{1 - (-1)}{1 - (-1)} = \frac{2}{2} = 1

f(x)=1x2f'(x) = -\frac{1}{x^2}

Setting 1c2=1-\frac{1}{c^2} = 1 gives c2=1c^2 = -1, which has no real solution!

This makes sense because the function is discontinuous - there's no point where the tangent slope equals the average slope.

Answer: MVT does not apply because ff is not continuous on [1,1][-1, 1] (discontinuity at x=0x = 0).

3Problem 3medium

Question:

A car travels 180 miles in 3 hours. Use the Mean Value Theorem to prove that at some time during the trip, the car was traveling exactly 60 mph.

💡 Show Solution

Step 1: Set up the problem

Let s(t)s(t) = position at time tt (in miles)

We know:

  • s(0)=0s(0) = 0 (starting position)
  • s(3)=180s(3) = 180 (ending position after 3 hours)

Step 2: Apply Mean Value Theorem

Assume s(t)s(t) is continuous on [0,3][0, 3] and differentiable on (0,3)(0, 3).

(Reasonable assumption for position function)

By MVT, there exists a time cc in (0,3)(0, 3) such that:

s(c)=s(3)s(0)30s'(c) = \frac{s(3) - s(0)}{3 - 0}


Step 3: Interpret

s(c)s'(c) is the instantaneous velocity at time cc

s(c)=18003=1803=60 mphs'(c) = \frac{180 - 0}{3} = \frac{180}{3} = 60 \text{ mph}


Conclusion

By the Mean Value Theorem, there exists a time cc (somewhere between 0 and 3 hours) where the instantaneous velocity was exactly 60 mph.


Real-world interpretation:

The average speed was 180 miles3 hours=60\frac{180 \text{ miles}}{3 \text{ hours}} = 60 mph.

MVT guarantees that at some moment, the speedometer read exactly 60 mph!

Answer: By the Mean Value Theorem, at some time cc in (0,3)(0, 3) hours, the car's instantaneous speed s(c)=60s'(c) = 60 mph.