Continuity and Differentiability Ex 5.8: Rolle's & Mean Value Theorem

Welcome! In this section, we'll explore two fundamental theorems of calculus that are central to NCERT Exercise 5.8: Rolle's Theorem and the Mean Value Theorem (MVT). These aren't just abstract rules; they provide a powerful connection between a function's average rate of change over an interval and its instantaneous rate of change at a specific point. Rolle's Theorem is a special case that tells us if a smooth curve starts and ends at the same height, its tangent must be horizontal somewhere in between. The Mean Value Theorem generalizes this, stating that for any smooth curve, there's a point where the tangent is parallel to the line connecting the endpoints. Mastering these theorems will deepen your understanding of derivatives and their geometric meaning, preparing you for more advanced calculus and typical board exam questions. Let's dive in!

Understanding Rolle's Theorem

Rolle's Theorem is a special case of the Mean Value Theorem. It provides specific conditions under which a function must have a point where its derivative is zero. Think of it as a guarantee of a 'flat spot' or a horizontal tangent on a graph.

The Theorem States:
Let f be a real-valued function defined on the closed interval [a, b]. If:

  1. f is continuous on the closed interval [a, b],
  2. f is differentiable on the open interval (a, b), and
  3. f(a) = f(b).

Then, there exists at least one number c in the open interval (a, b) such that f'(c) = 0.

Geometric Interpretation: Imagine a smooth, unbroken path from point A to point B. If point B is at the same elevation as point A, you must have traveled on a flat surface (where your instantaneous speed was zero relative to your starting height) at some moment during your journey. On a graph, this means if a continuous and smooth curve starts and ends at the same y-value, there must be at least one peak or valley between those points where the tangent line is horizontal.

Understanding the Mean Value Theorem (MVT)

The Mean Value Theorem (MVT) is a more general and incredibly powerful result. It relates the average rate of change of a function over an interval to its instantaneous rate of change at a point within that interval.

The Theorem States:
Let f be a real-valued function defined on the closed interval [a, b]. If:

  1. f is continuous on the closed interval [a, b], and
  2. f is differentiable on the open interval (a, b).

Then, there exists at least one number c in the open interval (a, b) such that:
f'(c) = (f(b) - f(a)) / (b - a)

Geometric Interpretation: The expression (f(b) - f(a)) / (b - a) is the slope of the secant line connecting the endpoints (a, f(a)) and (b, f(b)). The term f'(c) is the slope of the tangent line at the point c. The MVT guarantees that for any smooth, continuous curve, there is at least one point c where the tangent line is parallel to the secant line joining the endpoints. If you drive 120 km in 2 hours, your average speed is 60 km/h. The MVT guarantees that at some instant during your trip, your car's speedometer must have read exactly 60 km/h.

How to Verify Rolle's and Mean Value Theorems

  1. Step 1: Check Continuity — Verify that the function f(x) is continuous on the closed interval [a, b]. Polynomials, sine, cosine, and exponential functions are continuous everywhere. For rational functions, ensure the denominator is not zero in the interval.
  2. Step 2: Check Differentiability — Verify that the function f(x) is differentiable on the open interval (a, b). The same functions mentioned above are also differentiable everywhere. Be cautious with functions involving absolute values, fractional powers, or greatest integer functions, as they may have non-differentiable points.
  3. Step 3: Check Endpoint Values (for Rolle's Theorem only) — If you are applying Rolle's Theorem, calculate f(a) and f(b). If f(a) = f(b), the theorem is applicable. If not, you cannot use Rolle's Theorem (but MVT might still apply).
  4. Step 4: Find the Derivative and Solve for 'c' — Calculate the derivative f'(x). - For Rolle's Theorem, set f'(c) = 0 and solve for c. - For the Mean Value Theorem, first calculate the slope of the secant line m = (f(b) - f(a)) / (b - a). Then, set f'(c) = m and solve for c.
  5. Step 5: Verify 'c' — Check if the value(s) of c you found in the previous step lie within the open interval (a, b). Any value of c outside this interval is not a valid solution according to the theorems. The theorem guarantees at least one such c exists within the interval.

Worked Examples from Ex 5.8

  • Example 1 (Rolle's Theorem): Verify Rolle's Theorem for the function f(x) = x² + 2x - 8 on the interval x ∈ [-4, 2]. Step 1 & 2: Check Continuity and Differentiability f(x) is a polynomial function, so it is continuous and differentiable for all real numbers. Therefore, it is continuous on [-4, 2] and differentiable on (-4, 2). Step 3: Check Endpoint Values f(-4) = (-4)² + 2(-4) - 8 = 16 - 8 - 8 = 0 f(2) = (2)² + 2(2) - 8 = 4 + 4 - 8 = 0 Since f(-4) = f(2), all three conditions of Rolle's Theorem are satisfied. Step 4: Find Derivative and Solve for 'c' f'(x) = 2x + 2 Set f'(c) = 0 => 2c + 2 = 0 => 2c = -2 => c = -1. Step 5: Verify 'c' The value c = -1 lies in the open interval (-4, 2). Thus, Rolle's Theorem is verified.
  • Example 2 (Mean Value Theorem): Verify Mean Value Theorem for the function f(x) = x³ - 5x² - 3x in the interval [1, 3]. Step 1 & 2: Check Continuity and Differentiability f(x) is a polynomial function, hence it is continuous on [1, 3] and differentiable on (1, 3). The conditions for MVT are satisfied. Step 3 (not needed for MVT), we proceed to find endpoint values for the slope. f(1) = (1)³ - 5(1)² - 3(1) = 1 - 5 - 3 = -7 f(3) = (3)³ - 5(3)² - 3(3) = 27 - 45 - 9 = -27 Step 4: Find Derivative and Solve for 'c' First, find the slope of the secant line: m = (f(3) - f(1)) / (3 - 1) = (-27 - (-7)) / 2 = -20 / 2 = -10 Next, find the derivative: f'(x) = 3x² - 10x - 3 Set f'(c) = -10: 3c² - 10c - 3 = -10 3c² - 10c + 7 = 0 Using the quadratic formula c = [-b ± sqrt(b²-4ac)]/2a or factoring: (3c - 7)(c - 1) = 0 This gives c = 7/3 or c = 1. Step 5: Verify 'c' We need a value of c in the open interval (1, 3). c = 1 is an endpoint, so we discard it. c = 7/3 ≈ 2.33, which lies in (1, 3). Thus, MVT is verified with c = 7/3.

Exam Traps and Key Points

Here are some crucial points to remember for your exams:

  • Don't Skip the Conditions: Many students jump straight to finding f'(c). Always state and check the continuity and differentiability conditions first. For polynomials, you can simply state they are continuous and differentiable everywhere. For other functions, be more careful.
  • Open vs. Closed Intervals: Continuity is required on the closed interval [a, b], but differentiability is only required on the open interval (a, b). c must also lie in the open interval (a, b). A common trap is finding a value for c that equals a or b; this value is not a valid solution.
  • Non-Applicability: Be prepared for questions where a theorem is not applicable. For example, f(x) = |x| is not differentiable at x=0, so Rolle's/MVT would not apply on an interval like [-1, 1]. Similarly, f(x) = tan(x) is not continuous on [0, π] because of the asymptote at x = π/2.
  • State the Conclusion Clearly: After finding a valid c, conclude your answer by stating, "Since c = [value] lies in (a, b), the theorem is verified."

Practice Questions with Solutions

  • Q: Verify Rolle's theorem for the function f(x) = x(x-3)² on the interval [0, 3]. A: Step 1: f(x) = x(x² - 6x + 9) = x³ - 6x² + 9x. As a polynomial, it is continuous on [0, 3] and differentiable on (0, 3). Step 2: Check endpoints: f(0) = 0(0-3)² = 0. f(3) = 3(3-3)² = 0. Since f(0) = f(3), Rolle's theorem is applicable. Step 3: Find the derivative: f'(x) = 3x² - 12x + 9. Step 4: Set f'(c) = 0: 3c² - 12c + 9 = 0 => c² - 4c + 3 = 0 => (c-1)(c-3) = 0. This gives c=1 or c=3. Step 5: Verify c. The value c must be in the open interval (0, 3). So, c=1 is the required value. c=3 is an endpoint and is discarded. Final answer: Rolle's theorem is verified with c=1.
  • Q: Using the Mean Value Theorem, find a point c on the curve y = x² - 2x for x in [1, 3] where the tangent is parallel to the secant line joining the endpoints. A: Step 1: f(x) = x² - 2x is a polynomial, so it is continuous on [1, 3] and differentiable on (1, 3). MVT is applicable. Step 2: Find endpoint values: f(1) = 1² - 2(1) = -1. f(3) = 3² - 2(3) = 9 - 6 = 3. Step 3: Calculate the slope of the secant line: m = (f(3) - f(1)) / (3 - 1) = (3 - (-1)) / 2 = 4 / 2 = 2. Step 4: Find the derivative: f'(x) = 2x - 2. Set f'(c) = m: 2c - 2 = 2 => 2c = 4 => c = 2. Step 5: Verify c. The value c=2 lies in the open interval (1, 3). Final answer: The point is c=2.
  • Q: Discuss the applicability of Rolle's Theorem for f(x) = |x| on [-1, 1]. A: Step 1: Check continuity. f(x) = |x| is continuous on [-1, 1]. The graph has no breaks. Step 2: Check endpoint values. f(-1) = |-1| = 1. f(1) = |1| = 1. So, f(-1) = f(1). Step 3: Check differentiability on (-1, 1). The function f(x) = |x| is not differentiable at x = 0, which is in the interval (-1, 1). The graph has a sharp corner at x=0. Step 4: Since one of the conditions (differentiability on the open interval) is not met, Rolle's Theorem is not applicable. Final answer: Rolle's Theorem is not applicable because the function is not differentiable at x=0 in (-1, 1).
  • Q: Verify Mean Value Theorem for f(x) = x² in the interval [2, 4]. A: Step 1: f(x) = x² is a polynomial, so it is continuous on [2, 4] and differentiable on (2, 4). MVT is applicable. Step 2: Find endpoint values: f(2) = 2² = 4. f(4) = 4² = 16. Step 3: Calculate the slope of the secant line: m = (f(4) - f(2)) / (4 - 2) = (16 - 4) / 2 = 12 / 2 = 6. Step 4: Find the derivative: f'(x) = 2x. Set f'(c) = m: 2c = 6 => c = 3. Step 5: Verify c. The value c=3 lies in the open interval (2, 4). Final answer: The Mean Value Theorem is verified with c=3.

Frequently Asked Questions

What is the main difference between Rolle's Theorem and the Mean Value Theorem (MVT)?

The Mean Value Theorem is a generalization of Rolle's Theorem. Rolle's Theorem is a special case of MVT that applies only when the function's values at the endpoints are equal, i.e., `f(a) = f(b)`. This condition leads to the conclusion `f'(c) = 0`.

Why does differentiability need to be on the open interval (a, b) and not the closed interval [a, b]?

Derivatives are defined by limits, which require checking the function's behavior on both sides of a point. At an endpoint like 'a', we can only check the limit from the right side. The theorems' proofs work by analyzing the function's behavior *inside* the interval, so differentiability at the endpoints isn't necessary.

Can there be more than one value for 'c' that satisfies the theorems?

Yes, absolutely. Both theorems guarantee the existence of *at least one* such point `c`. For a wavy function like `sin(x)`, there can be multiple points within an interval where the tangent is horizontal (for Rolle's) or parallel to the secant (for MVT).

What happens if a function fails one of the conditions?

If any of the conditions (continuity on `[a, b]` or differentiability on `(a, b)`) are not met, the theorem is not applicable. This means we cannot guarantee the existence of a point `c` with the stated property, though one might still exist by chance.