Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.3

Question 1. Explain why Rolle's theorem is not applicable to the following functions in the respective intervals.
(i) \( f(x) = \left| \frac{1}{x} \right|, x \in [-1, 1] \)
(ii) \( f(x) = \tan x, x \in [0, \pi] \)
(iii) \( f(x) = x - 2 \log x, x \in [2, 7] \)
Answer:
(i) For the function \( f(x) = \left| \frac{1}{x} \right| \), within the interval \( [-1, 1] \), the function is not defined at \( x=0 \). Therefore, it is not continuous at \( x=0 \). Since a function must be continuous over the closed interval to apply Rolle's theorem, Rolle's theorem cannot be used here.
(ii) For the function \( f(x) = \tan x \), within the interval \( [0, \pi] \), the function is not continuous at \( x = \frac{\pi}{2} \) because \( \tan(\frac{\pi}{2}) \) is undefined. Since continuity over the entire closed interval is a condition for Rolle's theorem, it is not applicable.
(iii) For the function \( f(x) = x - 2 \log x \), within the interval \( [2, 7] \):
First, calculate the function values at the endpoints:
\( f(2) = 2 - 2 \log 2 = 2 - \log 4 \)
\( f(7) = 7 - 2 \log 7 = 7 - \log 49 \)
Comparing these values, we see that \( f(2) \ne f(7) \). Rolle's theorem requires that the function values at the endpoints of the interval must be equal. Since this condition is not met, Rolle's theorem is not applicable.
In simple words: Rolle's theorem has three main rules: the function must be smooth (differentiable), unbroken (continuous), and have the same value at both ends of the interval. If any of these rules are broken, like being undefined, having a break, or different values at the ends, then we cannot use Rolle's theorem.

🎯 Exam Tip: Always check all three conditions for Rolle's Theorem: continuity, differentiability, and \( f(a) = f(b) \). If even one fails, the theorem doesn't apply.

Question 2. Using Rolle's theorem, determine the values of x at which the tangent is parallel to the x-axis for the following functions:
(i) \( f(x) = x^2 - x, x \in [0, 1] \)
(ii) \( f(x) = \frac{x^2-2x}{x+2}, x \in [-1, 6] \)
(iii) \( f(x) = \sqrt{x} - \frac{x}{3}, x \in [0, 9] \)
Answer:
(i) For \( f(x) = x^2 - x \), in the interval \( [0, 1] \):
First, check the endpoint values:
\( f(0) = 0^2 - 0 = 0 \)
\( f(1) = 1^2 - 1 = 0 \)
\( \implies f(0) = f(1) = 0 \). The condition for Rolle's theorem holds.
The function \( f(x) \) is a polynomial, so it is continuous on \( [0, 1] \) and differentiable on \( (0, 1) \).
Now, find the derivative \( f'(x) \):
\( f'(x) = 2x - 1 \)
For the tangent to be parallel to the x-axis, the derivative must be zero:
\( f'(x) = 0 \implies 2x - 1 = 0 \)
\( 2x = 1 \)
\( x = \frac{1}{2} \)
Since \( x = \frac{1}{2} \in (0, 1) \), this is the value where the tangent is parallel to the x-axis.

(ii) For \( f(x) = \frac{x^2-2x}{x+2} \), in the interval \( [-1, 6] \):
First, check the endpoint values:
\( f(-1) = \frac{(-1)^2 - 2(-1)}{-1+2} = \frac{1+2}{1} = 3 \)
\( f(6) = \frac{6^2 - 2(6)}{6+2} = \frac{36-12}{8} = \frac{24}{8} = 3 \)
\( \implies f(-1) = f(6) = 3 \). The condition for Rolle's theorem holds.
The function \( f(x) \) is a rational function, continuous on \( [-1, 6] \) (since \( x+2 \ne 0 \) for \( x \in [-1, 6] \)) and differentiable on \( (-1, 6) \).
Now, find the derivative \( f'(x) \) using the quotient rule:
\( f'(x) = \frac{(x+2)(2x-2) - (x^2-2x)(1)}{(x+2)^2} \)
\( = \frac{(2x^2+4x-2x-4) - (x^2-2x)}{(x+2)^2} \)
\( = \frac{2x^2+2x-4 - x^2+2x}{(x+2)^2} \)
\( = \frac{x^2+4x-4}{(x+2)^2} \)
For the tangent to be parallel to the x-axis, \( f'(x) = 0 \):
\( \frac{x^2+4x-4}{(x+2)^2} = 0 \)
\( \implies x^2+4x-4 = 0 \)
Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \):
\( x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-4)}}{2(1)} \)
\( x = \frac{-4 \pm \sqrt{16+16}}{2} \)
\( x = \frac{-4 \pm \sqrt{32}}{2} \)
\( x = \frac{-4 \pm 4\sqrt{2}}{2} \)
\( x = -2 \pm 2\sqrt{2} \)
Now check which of these values lies in the interval \( (-1, 6) \):
\( -2 + 2\sqrt{2} \approx -2 + 2(1.414) = -2 + 2.828 = 0.828 \). This value is in \( (-1, 6) \).
\( -2 - 2\sqrt{2} \approx -2 - 2.828 = -4.828 \). This value is not in \( (-1, 6) \).
So, \( x = -2 + 2\sqrt{2} \) is the value where the tangent is parallel to the x-axis.

(iii) For \( f(x) = \sqrt{x} - \frac{x}{3} \), in the interval \( [0, 9] \):
First, check the endpoint values:
\( f(0) = \sqrt{0} - \frac{0}{3} = 0 - 0 = 0 \)
\( f(9) = \sqrt{9} - \frac{9}{3} = 3 - 3 = 0 \)
\( \implies f(0) = f(9) = 0 \). The condition for Rolle's theorem holds.
The function \( f(x) \) is continuous on \( [0, 9] \) and differentiable on \( (0, 9) \).
Now, find the derivative \( f'(x) \):
\( f'(x) = \frac{d}{dx}(\sqrt{x}) - \frac{d}{dx}(\frac{x}{3}) = \frac{1}{2\sqrt{x}} - \frac{1}{3} \)
For the tangent to be parallel to the x-axis, \( f'(x) = 0 \):
\( \frac{1}{2\sqrt{x}} - \frac{1}{3} = 0 \)
\( \implies \frac{1}{2\sqrt{x}} = \frac{1}{3} \)
\( \implies 2\sqrt{x} = 3 \)
\( \implies \sqrt{x} = \frac{3}{2} \)
Square both sides:
\( x = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \)
Since \( x = \frac{9}{4} \in (0, 9) \), this is the value where the tangent is parallel to the x-axis.
In simple words: To find where the tangent is flat (parallel to the x-axis), we first make sure Rolle's theorem can be used. This means the function must be smooth, unbroken, and have the same height at the start and end points. Then, we find the slope of the tangent line by taking the derivative and set it to zero. The x-value we find from this is where the tangent is flat.

🎯 Exam Tip: Remember that the existence of such a point \( c \) for Rolle's theorem is guaranteed only if all its conditions are met. Always show your checks for continuity and differentiability.

Question 3. Explain why Lagrange's mean value theorem is not applicable to the following functions in the respective intervals:
(i) \( f(x) = \frac{1}{2\sqrt{x}}, x \in [-1, 2] \)
(ii) \( f(x) = |3x+1|, x \in [-1, 3] \)
Answer:
(i) For the function \( f(x) = \frac{1}{2\sqrt{x}} \), in the interval \( [-1, 2] \):
The term \( \sqrt{x} \) requires \( x \ge 0 \). Also, the function is undefined when \( \sqrt{x} = 0 \), i.e., at \( x=0 \). Since \( x=0 \) is within the interval \( [-1, 2] \), the function is not continuous at \( x=0 \). Lagrange's Mean Value Theorem (LMVT) requires the function to be continuous on the closed interval. Because this condition is violated, LMVT is not applicable.

(ii) For the function \( f(x) = |3x+1| \), in the interval \( [-1, 3] \):
The function \( f(x) = |3x+1| \) is continuous everywhere, so it is continuous on \( [-1, 3] \).
However, LMVT also requires the function to be differentiable on the open interval \( (-1, 3) \). An absolute value function is not differentiable where its argument is zero. In this case, \( 3x+1 = 0 \implies x = -\frac{1}{3} \).
Since \( x = -\frac{1}{3} \) is within the open interval \( (-1, 3) \), the function is not differentiable at this point. Therefore, LMVT is not applicable in this given interval.
In simple words: Lagrange's Mean Value Theorem says that if a function is unbroken and smooth over an interval, there is a point where the tangent line is parallel to the line connecting the endpoints. If the function is broken (not continuous) or has a sharp corner (not differentiable) within that interval, the theorem cannot be used.

🎯 Exam Tip: For absolute value functions like \( |ax+b| \), always check differentiability at \( x = -b/a \). For functions with square roots or denominators, check where they become undefined or zero.

Question 4. Using the Lagrange's mean value theorem determine the values of x at which the tangent is parallel to the secant line at the end points of the given interval:
(i) \( f(x) = x^3 - 3x + 2, x \in [-2, 2] \)
(ii) \( f(x) = (x-2)(x-7), x \in [3, 11] \)
Answer:
(i) For \( f(x) = x^3 - 3x + 2 \), in the interval \( [-2, 2] \):
The function \( f(x) \) is a polynomial, so it is continuous on \( [-2, 2] \) and differentiable on \( (-2, 2) \).
First, find the values of the function at the endpoints:
\( f(-2) = (-2)^3 - 3(-2) + 2 = -8 + 6 + 2 = 0 \)
\( f(2) = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4 \)
Next, find the derivative \( f'(x) \):
\( f'(x) = 3x^2 - 3 \)
According to Lagrange's Mean Value Theorem, there exists a point \( c \in (-2, 2) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute the values:
\( 3c^2 - 3 = \frac{f(2)-f(-2)}{2-(-2)} \)
\( 3c^2 - 3 = \frac{4-0}{2+2} \)
\( 3c^2 - 3 = \frac{4}{4} \)
\( 3c^2 - 3 = 1 \)
\( 3c^2 = 4 \)
\( c^2 = \frac{4}{3} \)
\( c = \pm \sqrt{\frac{4}{3}} = \pm \frac{2}{\sqrt{3}} \)
Both \( \frac{2}{\sqrt{3}} \approx \frac{2}{1.732} \approx 1.155 \) and \( -\frac{2}{\sqrt{3}} \approx -1.155 \) lie within the interval \( (-2, 2) \).
So, \( x = \pm \frac{2}{\sqrt{3}} \) are the values where the tangent is parallel to the secant line.

(ii) For \( f(x) = (x-2)(x-7) \), in the interval \( [3, 11] \):
The function \( f(x) \) is a polynomial \( (x^2 - 9x + 14) \), so it is continuous on \( [3, 11] \) and differentiable on \( (3, 11) \).
First, find the values of the function at the endpoints:
\( f(3) = (3-2)(3-7) = (1)(-4) = -4 \)
\( f(11) = (11-2)(11-7) = (9)(4) = 36 \)
Next, find the derivative \( f'(x) \):
\( f(x) = x^2 - 9x + 14 \)
\( f'(x) = 2x - 9 \)
According to Lagrange's Mean Value Theorem, there exists a point \( c \in (3, 11) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute the values:
\( 2c - 9 = \frac{f(11)-f(3)}{11-3} \)
\( 2c - 9 = \frac{36 - (-4)}{8} \)
\( 2c - 9 = \frac{40}{8} \)
\( 2c - 9 = 5 \)
\( 2c = 14 \)
\( c = 7 \)
Since \( c = 7 \in (3, 11) \), this is the value where the tangent is parallel to the secant line.
In simple words: To find where a tangent line is parallel to the line connecting the two ends of a curve, we use Lagrange's Mean Value Theorem. First, ensure the function is continuous and smooth. Then, we find the slope of the secant line (connecting the endpoints) and set it equal to the derivative of the function (the slope of the tangent). Solving this equation gives us the x-value where these slopes match.

🎯 Exam Tip: Always calculate \( \frac{f(b)-f(a)}{b-a} \) accurately and then equate it to \( f'(c) \). Ensure the obtained value of \( c \) lies within the open interval \( (a, b) \).

Question 5. Show that the value in the conclusion of the mean value theorem for
(i) \( f(x) = \frac{1}{x} \) on a closed interval of positive numbers \( [a, b] \) is \( \sqrt{ab} \)
(ii) \( f(x) = Ax^2 + Bx + C \) on any interval \( [a, b] \) is \( \frac{a+b}{2} \)
Answer:
(i) For \( f(x) = \frac{1}{x} \), on the interval \( [a, b] \) where \( a, b > 0 \):
The function \( f(x) = \frac{1}{x} \) is continuous on \( [a, b] \) and differentiable on \( (a, b) \) for positive \( x \).
First, find the derivative \( f'(x) \):
\( f'(x) = -\frac{1}{x^2} \)
According to Lagrange's Mean Value Theorem, there exists a point \( c \in (a, b) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute the values:
\[ -\frac{1}{c^2} = \frac{\frac{1}{b} - \frac{1}{a}}{b-a} \]
\[ -\frac{1}{c^2} = \frac{\frac{a-b}{ab}}{b-a} \]
\[ -\frac{1}{c^2} = \frac{-(b-a)}{ab(b-a)} \]
\[ -\frac{1}{c^2} = -\frac{1}{ab} \]
\[ c^2 = ab \]
\[ c = \pm \sqrt{ab} \]
Since the interval \( [a, b] \) contains only positive numbers, \( c \) must be positive. Therefore, \( c = \sqrt{ab} \).
We know that for positive \( a, b \), \( \sqrt{ab} \) always lies between \( a \) and \( b \), so \( \sqrt{ab} \in (a,b) \). This proves the statement.

(ii) For \( f(x) = Ax^2 + Bx + C \), on the interval \( [a, b] \):
The function \( f(x) \) is a polynomial, so it is continuous on \( [a, b] \) and differentiable on \( (a, b) \).
First, find the derivative \( f'(x) \):
\( f'(x) = 2Ax + B \)
Next, find the values of the function at the endpoints:
\( f(a) = Aa^2 + Ba + C \)
\( f(b) = Ab^2 + Bb + C \)
According to Lagrange's Mean Value Theorem, there exists a point \( c \in (a, b) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute the values:
\[ 2Ac + B = \frac{(Ab^2+Bb+C) - (Aa^2+Ba+C)}{b-a} \]
\[ 2Ac + B = \frac{Ab^2+Bb+C-Aa^2-Ba-C}{b-a} \]
\[ 2Ac + B = \frac{A(b^2-a^2) + B(b-a)}{b-a} \]
Recall that \( b^2-a^2 = (b-a)(b+a) \):
\[ 2Ac + B = \frac{A(b-a)(b+a) + B(b-a)}{b-a} \]
Factor out \( (b-a) \) from the numerator:
\[ 2Ac + B = \frac{(b-a)[A(b+a) + B]}{b-a} \]
\[ 2Ac + B = A(b+a) + B \]
Subtract \( B \) from both sides:
\[ 2Ac = A(b+a) \]
Assuming \( A \ne 0 \), divide by \( 2A \):
\[ c = \frac{a+b}{2} \]
The midpoint \( \frac{a+b}{2} \) always lies within the interval \( (a, b) \). Thus, the value in the conclusion of the mean value theorem is \( \frac{a+b}{2} \).
In simple words: This question shows how the Mean Value Theorem finds a special point. For \( f(x) = 1/x \), this point is the geometric mean \( \sqrt{ab} \). For \( f(x) = Ax^2+Bx+C \), which is a parabola, the special point is always exactly in the middle of the interval, which is the arithmetic mean \( (a+b)/2 \). These are specific examples of the theorem working out.

🎯 Exam Tip: For quadratic functions, the point \( c \) from LMVT is always the midpoint of the interval. For functions like \( 1/x \), remember to consider the domain restrictions for \( c \).

Question 6. A race car driver is racing at 20th km. If his speed never exceeds 150 km/hr, what is the maximum distance he can cover in the next two hours?
Answer:
Let \( f(t) \) represent the distance covered by the race car driver at time \( t \).
We are given the initial position at \( t=0 \) (the start of the "next two hours") as \( f(0) = 20 \) km.
The time interval is \( [0, 2] \) hours, so we have \( a=0 \) and \( b=2 \).
The speed of the car is \( f'(t) \). We are told his speed never exceeds 150 km/hr, which means \( f'(t) \le 150 \).
We need to find the maximum distance he can cover in the next two hours, which is \( f(2) \).
Using Lagrange's Mean Value Theorem, there exists a point \( c \in (0, 2) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
\[ f'(c) = \frac{f(2)-f(0)}{2-0} \]
We know that \( f'(c) \le 150 \), so:
\[ \frac{f(2)-f(0)}{2-0} \le 150 \]
Substitute \( f(0) = 20 \):
\[ \frac{f(2)-20}{2} \le 150 \]
Multiply both sides by 2:
\[ f(2)-20 \le 300 \]
Add 20 to both sides:
\[ f(2) \le 320 \]
Therefore, the maximum distance the driver can cover in the next two hours is 320 km.
In simple words: The car starts at 20 km and its speed is never more than 150 km per hour. To find the longest distance it can go in 2 hours, we use a math rule. This rule tells us that the total distance covered will be 20 km (start) plus the maximum distance it can travel at its top speed for 2 hours, which is \( 150 \times 2 = 300 \) km. So, the total maximum distance is \( 20 + 300 = 320 \) km.

🎯 Exam Tip: When dealing with maximum/minimum distance or value problems, Lagrange's Mean Value Theorem helps relate the average rate of change to the instantaneous rate of change (derivative) and establish bounds.

Question 7. Suppose that for a function f(x), \( f'(x) \le 1 \) for all \( 1 \le x \le 4 \). Show that \( f(4) - f(1) \le 3 \).
Answer:
We are given a function \( f(x) \) for which \( f'(x) \le 1 \) for all \( x \) in the interval \( [1, 4] \).
Let \( a = 1 \) and \( b = 4 \).
Assuming \( f(x) \) is continuous on \( [1, 4] \) and differentiable on \( (1, 4) \) (which is implied by the existence of \( f'(x) \)), we can apply Lagrange's Mean Value Theorem.
According to LMVT, there exists a point \( c \in (1, 4) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute the given interval endpoints:
\[ f'(c) = \frac{f(4)-f(1)}{4-1} \]
\[ f'(c) = \frac{f(4)-f(1)}{3} \]
We are given that \( f'(x) \le 1 \) for all \( x \in [1, 4] \). This means \( f'(c) \le 1 \).
So, we can write the inequality:
\[ \frac{f(4)-f(1)}{3} \le 1 \]
Multiply both sides by 3:
\[ f(4)-f(1) \le 3 \]
Hence, it is proved that \( f(4) - f(1) \le 3 \).
In simple words: If a function's slope is always 1 or less, then for every 1 unit you move to the right, the function's value can go up by at most 1 unit. Here, we move 3 units from \( x=1 \) to \( x=4 \). So, the total change in the function's value, \( f(4) - f(1) \), cannot be more than \( 1 \times 3 = 3 \).

🎯 Exam Tip: When asked to prove an inequality involving function values and a bound on the derivative, LMVT is often the key tool. Remember to explicitly state the theorem's application and conditions.

Question 8. Does there exist a differentiable function f(x) such that \( f(0) = -1, f(2) = 4 \) and \( f'(x) \le 2 \) for all x. Justify your answer.
Answer:
No, such a differentiable function \( f(x) \) does not exist.
We are given \( f(0) = -1 \) and \( f(2) = 4 \). The interval is \( [0, 2] \).
We are also given that \( f(x) \) is differentiable, which implies it is continuous over the interval \( [0, 2] \).
According to Lagrange's Mean Value Theorem, there must exist a point \( c \in (0, 2) \) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute the given values:
\[ f'(c) = \frac{f(2)-f(0)}{2-0} \]
\[ f'(c) = \frac{4-(-1)}{2} \]
\[ f'(c) = \frac{5}{2} = 2.5 \]
This means that at some point \( c \) in the interval \( (0, 2) \), the derivative \( f'(c) \) must be 2.5.
However, we are given the condition that \( f'(x) \le 2 \) for all \( x \). This creates a contradiction: \( 2.5 \le 2 \) is false.
Since the conclusion from LMVT (that \( f'(c) = 2.5 \)) contradicts the given condition (that \( f'(x) \le 2 \)), such a function cannot exist.
In simple words: Imagine a hill. If you start at -1 and climb to 4 over a distance of 2, your average climbing speed (slope) must be 2.5. But if someone says you can never climb faster than 2, then it's impossible to get from -1 to 4 in just 2 steps. The required average slope (2.5) is higher than the maximum allowed slope (2).

🎯 Exam Tip: Questions asking "Does there exist..." often involve proving non-existence by showing a contradiction with a theorem like LMVT or Rolle's. Calculate the average rate of change and compare it to the given derivative bounds.

Question 9. Show that there lies a point on the curve \( f(x) = x(x + 3)e^{-\pi/2}, -3 \le x \le 0 \) where tangent drawn is parallel to the x-axis.
Answer:
Given the function \( f(x) = x(x+3)e^{-\pi/2} \) on the interval \( [-3, 0] \).
First, we simplify the function: \( f(x) = (x^2+3x)e^{-\pi/2} \). Here, \( e^{-\pi/2} \) is a constant.
We need to check the conditions for Rolle's Theorem:
1. **Continuity:** Since \( f(x) \) is a product of a polynomial \( (x^2+3x) \) and a constant \( e^{-\pi/2} \), it is continuous on the closed interval \( [-3, 0] \).
2. **Differentiability:** \( f(x) \) is a polynomial multiplied by a constant, so it is differentiable on the open interval \( (-3, 0) \).
3. **Function values at endpoints:**
\( f(-3) = (-3)(-3+3)e^{-\pi/2} = (-3)(0)e^{-\pi/2} = 0 \)
\( f(0) = (0)(0+3)e^{-\pi/2} = (0)(3)e^{-\pi/2} = 0 \)
Since \( f(-3) = f(0) = 0 \), the third condition of Rolle's Theorem is satisfied.
Because all conditions of Rolle's Theorem are met, there exists at least one point \( c \in (-3, 0) \) such that \( f'(c) = 0 \).
The condition \( f'(c) = 0 \) means the tangent drawn to the curve at \( x=c \) is parallel to the x-axis.
To find this point \( c \), we first find the derivative \( f'(x) \):
\( f'(x) = \frac{d}{dx}((x^2+3x)e^{-\pi/2}) \)
\( f'(x) = e^{-\pi/2} \frac{d}{dx}(x^2+3x) \)
\( f'(x) = e^{-\pi/2}(2x+3) \)
Now, set \( f'(x) = 0 \):
\( e^{-\pi/2}(2x+3) = 0 \)
Since \( e^{-\pi/2} \) is a constant and never zero, we must have:
\( 2x+3 = 0 \)
\( 2x = -3 \)
\( x = -\frac{3}{2} \)
The value \( c = -\frac{3}{2} \) lies within the open interval \( (-3, 0) \). This proves that such a point exists on the curve where the tangent is parallel to the x-axis.
In simple words: We want to show there's a point on the curve where the slope is zero (meaning the tangent is flat). Rolle's Theorem tells us this will happen if the function is smooth, unbroken, and starts and ends at the same height. We checked all these, found them to be true, and then calculated the exact point where the slope is zero.

🎯 Exam Tip: For Rolle's Theorem, make sure to explicitly state and verify all three conditions: continuity, differentiability, and \( f(a) = f(b) \). Showing the existence of \( c \) by calculation is often required.

Question 10. Using Mean Value Theorem prove that for, \( a > 0, b > 0, |e^{-a} - e^{-b}| < |a-b| \)
Answer:
Let \( f(x) = e^{-x} \). We want to prove \( |f(a) - f(b)| < |a-b| \). This is equivalent to proving \( \left| \frac{f(b)-f(a)}{b-a} \right| < 1 \).
The function \( f(x) = e^{-x} \) is continuous and differentiable for all real numbers, and thus on any interval \( [a, b] \) where \( a, b > 0 \).
First, find the derivative \( f'(x) \):
\( f'(x) = -e^{-x} \)
According to Lagrange's Mean Value Theorem, there exists a point \( c \) between \( a \) and \( b \) (i.e., \( c \in (a,b) \)) such that:
\[ f'(c) = \frac{f(b)-f(a)}{b-a} \]
Substitute \( f'(c) = -e^{-c} \):
\[ -e^{-c} = \frac{e^{-b}-e^{-a}}{b-a} \]
Now, consider the absolute value of both sides:
\[ |-e^{-c}| = \left| \frac{e^{-b}-e^{-a}}{b-a} \right| \]
\[ e^{-c} = \left| \frac{e^{-b}-e^{-a}}{b-a} \right| \]
We know that \( a > 0 \) and \( b > 0 \). Since \( c \) is between \( a \) and \( b \), it must also be positive, i.e., \( c > 0 \).
For any positive value of \( c \), the exponential term \( e^{-c} \) is always less than 1 (because \( e^{-c} = 1/e^c \), and \( e^c > 1 \) for \( c > 0 \)).
So, \( e^{-c} < 1 \).
Therefore, we have:
\[ \left| \frac{e^{-b}-e^{-a}}{b-a} \right| < 1 \]
Multiplying both sides by \( |b-a| \):
\[ |e^{-b}-e^{-a}| < |b-a| \]
This is the same as \( |e^{-a} - e^{-b}| < |a-b| \), just with the order of terms swapped in the absolute value, which does not change its value. Hence, the statement is proved.
In simple words: We use the Mean Value Theorem to show that the change in \( e^{-x} \) over an interval is always smaller than the length of that interval. This is because the slope of \( e^{-x} \) (which is \( -e^{-x} \)) always has a value between 0 and 1, meaning it is never steeper than 1. So, the "average steepness" is also less than 1, leading to the given inequality.

🎯 Exam Tip: When proving inequalities using LMVT, remember to evaluate the derivative's bounds. The core idea is that \( |f'(c)| \) provides a direct bound for \( \left| \frac{f(b)-f(a)}{b-a} \right| \).