S.M. Yusuf Calculus Chapter 2 Mean Value Theorems Solved Exercises & Proofs PDF Guide | BS Math, ADP, Punjab University, UOS, QAU, FBISE Notes

Theoretical Foundations & Core Concepts

Mean Value Theorems form the bedrock of real analysis and differential calculus. In S.M. Yusuf's Calculus with Analytic Geometry, these theorems bridge the local behavior of a function (its derivative at a specific point) with its global behavior across a given interval.

1. Rolle's Theorem

Let $f: [a, b] \to \mathbb{R}$ be a real-valued function satisfying three fundamental conditions:

• Continuity: $f(x)$ is continuous on the closed interval $[a, b]$.
• Differentiability: $f(x)$ is differentiable on the open interval $(a, b)$.
• Equal Endpoints: $f(a) = f(b)$.

Then, there exists at least one point $c \in (a, b)$ such that $f'(c) = 0$. Geometrically, this guarantees at least one point on the curve where the tangent line is horizontal (parallel to the x-axis).

2. Lagrange's Mean Value Theorem (First Mean Value Theorem)

Lagrange's Mean Value Theorem (LMVT) relaxes the third condition of Rolle's Theorem. If $f: [a, b] \to \mathbb{R}$ satisfies:

• $f(x)$ is continuous on $[a, b]$.
• $f(x)$ is differentiable on $(a, b)$.

Then, there exists at least one point $c \in (a, b)$ such that:

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

Alternative Form: Setting $b = a + h$, where $h > 0$, any point $c \in (a, a+h)$ can be written as $c = a + \theta h$ where $0 < \theta < 1$. The theorem then assumes the form:

$$f(a + h) = f(a) + h \, f'(a + \theta h), \quad 0 < \theta < 1$$

Geometrically, $f'(c)$ represents the slope of the tangent at $x = c$, which equals the slope of the secant line joining the endpoints $(a, f(a))$ and $(b, f(b))$.

3. Cauchy's Mean Value Theorem (Extended Mean Value Theorem)

Cauchy's Mean Value Theorem generalizes LMVT to two functions $f(x)$ and $g(x)$. Let $f, g: [a, b] \to \mathbb{R}$ be two functions such that:

• Both $f(x)$ and $g(x)$ are continuous on $[a, b]$.
• Both $f(x)$ and $g(x)$ are differentiable on $(a, b)$.
• $g'(x) \neq 0$ for all $x \in (a, b)$.

Then, there exists at least one point $c \in (a, b)$ such that:

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

Note: When $g(x) = x$, Cauchy's Mean Value Theorem reduces directly to Lagrange's Mean Value Theorem.

Formula Summary & Quick Reference

Theorem	Hypotheses	Key Formula / Relation	Geometric Interpretation
Rolle's Theorem	Cont. on $[a,b]$, Diff. on $(a,b)$, $f(a)=f(b)$	$f'(c) = 0$ for $c \in (a,b)$	Horizontal tangent exists at $x = c$.
Lagrange's MVT	Cont. on $[a,b]$, Diff. on $(a,b)$	$f'(c) = \frac{f(b)-f(a)}{b-a}$	Tangent at $c$ is parallel to the secant line.
LMVT ($\theta$-Form)	Cont. on $[a, a+h]$, Diff. on $(a, a+h)$	$f(a+h) = f(a) + h \, f'(a+\theta h)$, $0 < \theta < 1$	Parametric expression for interior point $c$.
Cauchy's MVT	$f, g$ Cont. on $[a,b]$, Diff. on $(a,b)$, $g'(x) \neq 0$	$\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$	Ratio of instantaneous rates equals ratio of net changes.

Step-by-Step Solved Board Exam Questions

Question 1 (Punjab University Past Paper)

Problem: Verify Rolle's Theorem for $f(x) = x^3 - 4x$ on the interval $[-2, 2]$ and find all values of $c \in (-2, 2)$ that satisfy the theorem.

Solution:

Step 1: Check Continuity and Differentiability.
$f(x) = x^3 - 4x$ is a polynomial function. Polynomials are continuous and differentiable everywhere on $\mathbb{R}$. Thus:

• $f(x)$ is continuous on $[-2, 2]$.
• $f(x)$ is differentiable on $(-2, 2)$.

Step 2: Check Endpoint Values.
Evaluate $f(-2)$ and $f(2)$:

$$f(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0$$ $$f(2) = (2)^3 - 4(2) = 8 - 8 = 0$$

Since $f(-2) = f(2) = 0$, all three hypotheses of Rolle's Theorem are satisfied.

Step 3: Find $c \in (-2, 2)$ such that $f'(c) = 0$.
Compute the derivative:

$$f'(x) = 3x^2 - 4$$

Set $f'(c) = 0$:

$$3c^2 - 4 = 0 \implies c^2 = \frac{4}{3} \implies c = \pm \frac{2}{\sqrt{3}}$$

Since $\frac{2}{\sqrt{3}} \approx 1.155$, both values $c_1 = \frac{2}{\sqrt{3}}$ and $c_2 = -\frac{2}{\sqrt{3}}$ lie strictly within the open interval $(-2, 2)$. Hence, Rolle's Theorem is verified.

---

Question 2 (University of Sargodha Past Paper)

Problem: Apply Lagrange's Mean Value Theorem to $f(x) = \sqrt{x}$ on $[1, 4]$ and find the value of $c$. Also find the value of $\theta$ using the form $f(a+h) = f(a) + h \, f'(a+\theta h)$.

Solution:

Step 1: Verify Hypotheses.
$f(x) = \sqrt{x}$ is continuous on $[1, 4]$ and differentiable on $(1, 4)$ with derivative $f'(x) = \frac{1}{2\sqrt{x}}$.

Step 2: Apply LMVT Formula.
By LMVT, there exists $c \in (1, 4)$ such that:

$$f'(c) = \frac{f(4) - f(1)}{4 - 1}$$ $$\frac{1}{2\sqrt{c}} = \frac{\sqrt{4} - \sqrt{1}}{3} = \frac{2 - 1}{3} = \frac{1}{3}$$ $$\frac{1}{2\sqrt{c}} = \frac{1}{3} \implies 2\sqrt{c} = 3 \implies \sqrt{c} = \frac{3}{2} \implies c = \frac{9}{4} = 2.25$$

Since $2.25 \in (1, 4)$, LMVT is verified with $c = \frac{9}{4}$.

Step 3: Calculate $\theta$.
Here $a = 1$, $h = b - a = 4 - 1 = 3$. We know $c = a + \theta h$:

$$\frac{9}{4} = 1 + \theta(3) \implies 3\theta = \frac{9}{4} - 1 = \frac{5}{4} \implies \theta = \frac{5}{12}$$

Notice that $0 < \theta = \frac{5}{12} < 1$, as required by the theory.

---

Question 3 (Quaid-i-Azam University Past Paper)

Problem: Verify Cauchy's Mean Value Theorem for $f(x) = x^2$ and $g(x) = x^3$ on $[1, 2]$.

Solution:

Step 1: Check Conditions.
$f(x) = x^2$ and $g(x) = x^3$ are polynomials, so they are continuous on $[1, 2]$ and differentiable on $(1, 2)$.
Derivatives: $f'(x) = 2x$, $g'(x) = 3x^2$.
On $(1, 2)$, $g'(x) = 3x^2 \neq 0$. Thus, all conditions are met.

Step 2: Apply Cauchy's MVT Formula.
There exists $c \in (1, 2)$ such that:

$$\frac{f'(c)}{g'(c)} = \frac{f(2) - f(1)}{g(2) - g(1)}$$ $$\frac{2c}{3c^2} = \frac{2^2 - 1^2}{2^3 - 1^3}$$ $$\frac{2}{3c} = \frac{4 - 1}{8 - 1} = \frac{3}{7}$$

Step 3: Solve for $c$.

$$\frac{2}{3c} = \frac{3}{7} \implies 9c = 14 \implies c = \frac{14}{9} \approx 1.556$$

Since $1.556 \in (1, 2)$, Cauchy's Mean Value Theorem is verified.

---

Question 4 (FBISE / University Annual Exam Question)

Problem: Use Lagrange's Mean Value Theorem to prove that for $0 < a < b$:

$$\frac{b - a}{1 + b^2} < \tan^{-1}b - \tan^{-1}a < \frac{b - a}{1 + a^2}$$

Solution:

Step 1: Define Function and Interval.
Let $f(x) = \tan^{-1}x$ on the interval $[a, b]$, where $0 < a < b$.

Step 2: Apply LMVT.
$f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$ with $f'(x) = \frac{1}{1 + x^2}$.
By LMVT, there exists $c \in (a, b)$ such that:

$$f'(c) = \frac{f(b) - f(a)}{b - a} \implies \frac{1}{1 + c^2} = \frac{\tan^{-1}b - \tan^{-1}a}{b - a}$$

Step 3: Establish Bounds for $c$.
Since $a < c < b$, squaring and adding 1 preserves order for positive quantities:

$$a < c < b \implies a^2 < c^2 < b^2 \implies 1 + a^2 < 1 + c^2 < 1 + b^2$$

Taking reciprocals reverses the inequalities:

$$\frac{1}{1 + b^2} < \frac{1}{1 + c^2} < \frac{1}{1 + a^2}$$

Step 4: Substitute and Multiply.
Substitute $\frac{1}{1 + c^2} = \frac{\tan^{-1}b - \tan^{-1}a}{b - a}$ into the inequality:

$$\frac{1}{1 + b^2} < \frac{\tan^{-1}b - \tan^{-1}a}{b - a} < \frac{1}{1 + a^2}$$

Since $b - a > 0$, multiply throughout by $(b - a)$:

$$\frac{b - a}{1 + b^2} < \tan^{-1}b - \tan^{-1}a < \frac{b - a}{1 + a^2}$$

This completes the proof.

Interactive Practice Quiz (Clickable MCQs)

Frequently Asked Questions (FAQs)

1. Why is continuity required on the closed interval $[a,b]$, but differentiability only on the open interval $(a,b)$?

Continuity is required on the closed interval $[a, b]$ to guarantee that the extreme values (maximum and minimum) are actually attained by the function by the Extreme Value Theorem. Differentiability is only required on the open interval $(a, b)$ because derivatives at endpoints require one-sided limits; interior points are sufficient to establish the existence of a horizontal tangent or parallel secant line.

2. How do Mean Value Theorems help prove mathematical inequalities in university exams?

In university examinations (such as PU, UOS, BZU, and QAU), LMVT is a standard tool for establishing inequalities. By choosing an appropriate function $f(x)$ on an interval $[a, b]$, applying $f'(c) = \frac{f(b)-f(a)}{b-a}$, and bounding $f'(c)$ between its minimum and maximum values over $(a, b)$, one directly produces lower and upper bounds for $f(b) - f(a)$.

3. What is the key distinction between Lagrange's MVT and Cauchy's MVT?

Lagrange's MVT applies to a single function $f(x)$ and compares its average rate of change to its instantaneous derivative. Cauchy's MVT applies to two functions $f(x)$ and $g(x)$ simultaneously, comparing the ratio of their instantaneous rates of change to the ratio of their overall changes over the interval $[a, b]$. Cauchy's MVT is particularly crucial for proving L'Hôpital's Rule.