Unit 1.16 – Working with the Intermediate Value Theorem (IVT)

THE 5-STEP IVT APPLICATION PROCESS

1. STEP 1: Verify Continuity
   • Check that \(f(x)\) is continuous on \([a, b]\)
   • Look for: polynomials (always continuous), rational functions (check denominator ≠ 0), trig functions (continuous on domain), piecewise (check boundaries)
   • If discontinuous anywhere in \([a, b]\) → STOP! Cannot use IVT
2. STEP 2: Evaluate Endpoints
   • Calculate \(f(a)\) and \(f(b)\)
   • Make sure both values exist and are finite
3. STEP 3: Identify Target Value d
   • Determine what value you're looking for (often \(d = 0\) for roots)
   • Verify that \(d\) is between \(f(a)\) and \(f(b)\)
   • Check: \(f(a) \leq d \leq f(b)\) OR \(f(b) \leq d \leq f(a)\)
4. STEP 4: State the IVT Conclusion
   • Write: "By the Intermediate Value Theorem, there exists at least one \(c \in (a, b)\) such that \(f(c) = d\)"
   • Use proper mathematical language for full credit on AP® exams!
5. STEP 5: (Optional) Approximate c if requested
   • Use bisection method, sign analysis, or calculator to narrow down the location
   • IVT proves existence; finding the exact value requires additional techniques

Problem: Prove that \(f(x) = x^3 - 2x - 5\) has at least one root in the interval \([2, 3]\)

Solution:

1. Step 1 - Verify continuity:
   • \(f(x)\) is a polynomial
   • Polynomials are continuous everywhere ✓
2. Step 2 - Evaluate endpoints:
   • \(f(2) = 2^3 - 2(2) - 5 = 8 - 4 - 5 = -1\)
   • \(f(3) = 3^3 - 2(3) - 5 = 27 - 6 - 5 = 16\)
3. Step 3 - Check sign change:
   • \(f(2) = -1 < 0\) and \(f(3) = 16 > 0\)
   • Opposite signs! ✓
   • Therefore, \(0\) is between \(f(2)\) and \(f(3)\)
4. Step 4 - Apply IVT:
   • Since \(f\) is continuous on \([2, 3]\) and \(f(2) < 0 < f(3)\)
   • By the Intermediate Value Theorem, there exists at least one \(c \in (2, 3)\) such that \(f(c) = 0\)

Conclusion: The equation \(x^3 - 2x - 5 = 0\) has at least one solution in the interval \((2, 3)\) ✓

Problem: Let \(g(x) = x^2 + 2x\) on \([0, 4]\). Show that there exists a value \(c\) where \(g(c) = 10\)

Solution:

1. Continuity: \(g(x)\) is a polynomial, so continuous on \([0, 4]\) ✓
2. Evaluate endpoints:
   • \(g(0) = 0^2 + 2(0) = 0\)
   • \(g(4) = 4^2 + 2(4) = 16 + 8 = 24\)
3. Check if 10 is between endpoint values:
   • \(0 < 10 < 24\) ✓
   • Yes! \(10\) is between \(g(0)\) and \(g(4)\)
4. Apply IVT:
   • Since \(g\) is continuous on \([0, 4]\) and \(10\) lies between \(g(0) = 0\) and \(g(4) = 24\)
   • By IVT, there exists at least one \(c \in (0, 4)\) such that \(g(c) = 10\)

Note: We can actually solve algebraically: \(x^2 + 2x = 10 \Rightarrow x^2 + 2x - 10 = 0\) gives \(x = 2.162\) (in the interval)

Problem: Show that \(\sin(x) = \frac{1}{2}\) has a solution in \([0, \pi]\)

Solution:

1. Continuity: \(\sin(x)\) is continuous everywhere ✓
2. Evaluate endpoints:
   • \(\sin(0) = 0\)
   • \(\sin(\pi) = 0\)
3. Problem? Both endpoints equal 0, but we want \(\frac{1}{2}\)
   • However, we know \(\sin(x)\) reaches a maximum of 1 at \(x = \frac{\pi}{2}\)
   • Since \(\sin\) is continuous and \(\sin(\frac{\pi}{2}) = 1\)
4. Better approach - use different interval:
   • On \([0, \frac{\pi}{2}]\): \(\sin(0) = 0\) and \(\sin(\frac{\pi}{2}) = 1\)
   • Since \(0 < \frac{1}{2} < 1\), IVT guarantees a \(c \in (0, \frac{\pi}{2})\) where \(\sin(c) = \frac{1}{2}\)

Answer: Yes, \(c = \frac{\pi}{6}\) (we know this from the unit circle)

Problem: Can we use IVT to show \(h(x) = \frac{1}{x - 2}\) equals \(0\) somewhere on \([1, 3]\)?

Solution:

1. Check continuity:
   • \(h(x)\) has a vertical asymptote at \(x = 2\)
   • \(x = 2\) is in the interval \([1, 3]\)
   • Therefore, \(h(x)\) is NOT continuous on \([1, 3]\) ✗
2. Conclusion:
   • Cannot apply IVT!
   • The first condition (continuity) fails
   • Note: \(h(x)\) never equals 0 anyway (rational function with constant numerator)

Lesson: Always check continuity FIRST before trying to apply IVT!

Problem: A car's temperature gauge reads 60°F at 8:00 AM and 85°F at 10:00 AM. Assuming temperature varies continuously, prove there was a moment when the temperature was exactly 75°F.

Solution:

1. Define the function:
   • Let \(T(t)\) = temperature at time \(t\)
   • Interval: \([8, 10]\) (hours after midnight)
2. Check conditions:
   • Given: Temperature varies continuously ✓
   • \(T(8) = 60°F\)
   • \(T(10) = 85°F\)
   • Target: \(75°F\), and \(60 < 75 < 85\) ✓
3. Apply IVT:
   • Since \(T(t)\) is continuous on \([8, 10]\) and 75 is between \(T(8)\) and \(T(10)\)
   • By IVT, there exists at least one time \(c \in (8, 10)\) when \(T(c) = 75°F\)

Answer: Yes, the temperature was exactly 75°F at some moment between 8:00 AM and 10:00 AM

For each problem, determine if IVT can be applied and what it guarantees:

1. Does \(f(x) = x^3 + x - 1\) have a root in \([0, 1]\)?
2. Show that \(\cos(x) = x\) has a solution in \([0, 1]\)
3. Can IVT prove that \(g(x) = \frac{x^2 - 4}{x - 2}\) equals 3 on \([1, 3]\)?
4. A population grows from 100 to 500 continuously over 5 years. Was the population exactly 300 at some point?

Answers:

1. YES: \(f(0) = -1 < 0\), \(f(1) = 1 > 0\); sign change, continuous → root exists
2. YES: Define \(h(x) = \cos(x) - x\); \(h(0) = 1 > 0\), \(h(1) = \cos(1) - 1 < 0\); sign change → solution exists
3. NO: \(g(x)\) has removable discontinuity at \(x = 2\) (in the interval); not continuous on \([1, 3]\)
4. YES: Define \(P(t)\) = population at year \(t\); \(P(0) = 100\), \(P(5) = 500\), \(100 < 300 < 500\); IVT guarantees \(P(c) = 300\) for some \(c\)