Imagine a rollercoaster that starts and ends at the same height 🎢. At some point, the track must be perfectly flat (horizontal tangent)! That's Rolle's Theorem in action!
1️⃣ Continuous on [a, b]
2️⃣ Differentiable on (a, b)
3️⃣ f(a) = f(b)
Then ∃ c ∈ (a, b) where f'(c) = 0
Mean Value Theorem (MVT)
If you drive 120km in 2 hours, your average speed is 60km/h. MVT says at some instant, your speedometer must have shown exactly 60km/h! 🚗💨
1️⃣ Continuous on [a, b]
2️⃣ Differentiable on (a, b)
Then ∃ c ∈ (a, b) where f'(c) = (f(b)-f(a))/(b-a)
MVT is like Rolle's Theorem but for tilted graphs! It's Rolle's cool cousin that doesn't require f(a) = f(b). 😎
Exercise 7.3.1 - Rolle's Check
Explain why Rolle's theorem is not applicable to the following functions:
1f(x) = |1/x|, x ∈ [-1,1]
🔍 Continuity Check: At x=0, we hit a vertical asymptote! 🚀 The function shoots to infinity → Discontinuous!
✏️ Differentiability Check: Can't draw a smooth tangent at x=0 → Not differentiable!
Rolle's Conditions Failed: 2/3 conditions fail (continuity and differentiability). Even though f(-1)=f(1)=1, the theorem doesn't apply here!
No horizontal tangent guaranteed because the function breaks at x=0. It's like a rollercoaster with a broken track - no smooth ride! 🎢💥
2f(x) = tan x, x ∈ [0,π]
🔍 Continuity Check: At x=π/2, tan x goes crazy to ±∞ → Big discontinuity!
✏️ Differentiability Check: No smooth tangent possible at π/2 → Not differentiable there!
Even though f(0)=f(π)=0 (both touch the x-axis), the function misbehaves in the middle! Like a rollercoaster that suddenly teleports to infinity. 🎢➡️🚀
Watch out! The function is actually continuous and differentiable on (0,π/2) and (π/2,π), but the single point at π/2 ruins everything for Rolle's theorem!
3f(x) = x - 2 log x, x ∈ [2,7]
🔍 Continuity Check: The function is continuous on [2,7] since log x is defined for x > 0 → Continuous!
✏️ Differentiability Check: The derivative exists for all x > 0 → Differentiable!
Rolle's Condition Failed: While the function is continuous and differentiable, the endpoint values aren't equal. The theorem doesn't apply here!
The function keeps increasing throughout the interval - it never returns to the same y-value. No horizontal tangent guaranteed in this case!
Exercise 7.3.2 - Find Horizontal Tangents
Using Rolle's theorem, find where the tangent is parallel to the x-axis:
1f(x) = x² - x, x ∈ [0,1]
🎯 Rolle's Conditions:
✔️ Continuous parabola (smooth everywhere)
✔️ Differentiable everywhere
✔️ f(0)=0 and f(1)=0 → Same height!
🔧 Find f'(x): 2x - 1 (derivative of x² - x)
⚖️ Set f'(x)=0: 2x - 1 = 0 → x = 0.5
Found it! At x=0.5, the tangent is perfectly horizontal. The parabola bottoms out exactly halfway between 0 and 1! ⚖️
This makes sense because parabolas are symmetric! The vertex (lowest/highest point) is always midway between roots when they exist.
2f(x) = (x² - 2x)/(x + 2), x ∈ [-1,6]
🎯 Rolle's Conditions:
✔️ Continuous except at x=-2 (not in our interval)
✔️ Differentiable on (-1,6)
✔️ f(-1)=3 and f(6)=3 → Same height!
🔧 Find f'(x): Using quotient rule → [(2x-2)(x+2)-(x²-2x)(1)]/(x+2)² = (x²+4x-4)/(x+2)²
⚖️ Set f'(x)=0: x²+4x-4 = 0 → x = [-4 ± √(16+16)]/2 = -2 ± 2√2
📍 Check interval: Only x = -2 + 2√2 ≈ 0.828 is in (-1,6)
Solution: At x ≈ 0.828, the tangent is horizontal. The function curves back to the same y-value at x=6 after dipping down!
3f(x) = √x - x/3, x ∈ [0,9]
🎯 Rolle's Conditions:
✔️ Continuous on [0,9]
✔️ Differentiable on (0,9)
✔️ f(0)=0 and f(9)=0 → Same height!
🔧 Find f'(x): 1/(2√x) - 1/3
⚖️ Set f'(x)=0: 1/(2√x) = 1/3 → 2√x = 3 → √x = 1.5 → x = 2.25
Peak Point: At x=2.25, the function reaches its maximum value before descending back to 0 at x=9. The tangent is perfectly flat at the peak!
Exercise 7.3.3 - MVT Not Applicable
Explain why Lagrange's MVT is not applicable to these functions:
1f(x) = (x+1)/x, x ∈ [-1,2]
🔍 Continuity Check: At x=0, we have division by zero → Discontinuous!
✏️ Differentiability Check: Can't be differentiable where it's not continuous → Not differentiable at x=0!
MVT Conditions Failed: The function is neither continuous nor differentiable on [-1,2] because of the discontinuity at x=0.
The function has a vertical asymptote at x=0, creating an infinite discontinuity. MVT requires a smooth, unbroken curve between the endpoints!
2f(x) = |3x + 1|, x ∈ [-1,3]
🔍 Continuity Check: Absolute value functions are always → Continuous!
✏️ Differentiability Check: At x = -1/3 (where 3x+1=0), there's a sharp corner → Not differentiable at x=-1/3!
MVT Condition Failed: While continuous everywhere, the function isn't differentiable at x=-1/3. MVT requires differentiability on the entire open interval.
Absolute value functions create "sharp turns" where their argument equals zero. At these points, we can't draw a unique tangent line!
Exercise 7.3.4 - MVT Applications
Using MVT, find where the tangent is parallel to the secant line:
Perfect Midpoint! The point x=7 is exactly midway between the roots at x=2 and x=7. The parabola's symmetry makes this the point where the tangent matches the secant slope.
⚖️ Set f'(c) = slope: -1/c² = -1/(ab) → c² = ab → c = √(ab)
Geometric Mean! The point c is the geometric mean of a and b. For the hyperbola 1/x, this special point always satisfies MVT.
2f(x) = Ax² + Bx + C on [a,b], show MVT point c = (a+b)/2
🎯 MVT Application:
✔️ Polynomial → Continuous everywhere
✔️ Polynomial → Differentiable everywhere
📏 Calculate slope: [f(b)-f(a)]/(b-a) = [A(b²-a²)+B(b-a)]/(b-a) = A(a+b) + B
🔧 Find f'(x): 2Ax + B
⚖️ Set f'(c) = slope: 2Ac + B = A(a+b) + B → 2Ac = A(a+b) → c = (a+b)/2
Perfect Symmetry! For any quadratic function, the point satisfying MVT is always exactly midway between a and b. This reflects the symmetry of parabolas!
Exercise 7.3.6 - Race Car Challenge
A race car at km 20 with speed ≤ 150 km/hr. What's the max km in 2 hours?
🚗 Position Function: s(t) = kilometer stone at time t hours
Maximum reach: 320 km in 2 hours if driving at constant 150 km/hr. Any speed variations would mean lower distance!
This is why police can estimate your speed between two points without radar guns! MVT is used in speed cameras too! 🚔📸
Exercise 7.3.7 - Function Bound
For f(x) with f'(x) ≤ 1 on [1,4], show f(4) - f(1) ≤ 3
🎯 Apply MVT: There exists c ∈ (1,4) such that f'(c) = [f(4)-f(1)]/(4-1)
📏 Given f'(x) ≤ 1: [f(4)-f(1)]/3 ≤ 1
🧮 Rearrange: f(4) - f(1) ≤ 3
Slope Interpretation: The average slope between x=1 and x=4 can't exceed the maximum derivative. Since f'(x) never exceeds 1, the total change in f(x) can't exceed 3 over 3 units.
This is a practical application of MVT for bounding function values based on derivative information - super useful in calculus!
Exercise 7.3.8 - Differentiable Function
Does a differentiable function exist with f(0)=-1, f(2)=4 and f'(x)≤2 for all x?
🎯 Apply MVT: If such f exists, then ∃ c ∈ (0,2) with f'(c) = [f(2)-f(0)]/(2-0) = 5/2 = 2.5
⚠️ Contradiction: But f'(x) ≤ 2 for all x, yet we need f'(c) = 2.5
❌ Conclusion: No such function exists because it would require a derivative value exceeding the given bound
To go from -1 to 4 over 2 units requires an average slope of 2.5. If the slope never exceeds 2, this is impossible! The function can't climb fast enough.
Exercise 7.3.9 - Horizontal Tangent
Show there's a point on f(x) = x(x+3)e^(-x/2) in [-3,0] with horizontal tangent
🎯 Rolle's Theorem Conditions:
✔️ Continuous on [-3,0] (composition of continuous functions)
✔️ Differentiable on (-3,0)
✔️ f(-3) = 0 and f(0) = 0 → Same height!
🔍 Conclusion: By Rolle's Theorem, ∃ c ∈ (-3,0) where f'(c) = 0
The function starts and ends at 0, and is smooth in between. It must have at least one peak or valley where the tangent is horizontal!
We don't even need to calculate the derivative! Rolle's Theorem guarantees the existence of such a point based on the function's behavior at the endpoints.
🎯 Apply MVT: (f(b)-f(a))/(b-a) = f'(c) for some c between a and b
✏️ Find f'(x): -e⁻ˣ → |f'(c)| = e⁻ᶜ
💡 Since c > 0: e⁻ᶜ < e⁰ = 1
The slope between any two points is always less steep than the initial slope at x=0 because the function flattens out as x increases.
🧮 Final Steps:
|(e⁻ᵇ - e⁻ᵃ)/(b-a)| = e⁻ᶜ < 1
Multiply both sides by |b-a| → |e⁻ᵇ - e⁻ᵃ| < |b-a|
This shows exponential functions change more slowly than linear functions! The difference in y-values is always less than the difference in x-values. 📉