Question 79 of 82intermediate🔧 ApplyLong Answer5 marks

Let ff be a function defined on [a,b][a, b] such that f′(x)>0f'(x) > 0, for all x∈(a,b)x \in (a, b). Then prove that ff is an increasing function on (a,b)(a, b).

Correct Answer

Let x1,x2∈[a,b]x_1, x_2 \in [a, b] be two points such that x1<x2x_1 < x_2.

By the Mean Value Theorem, there exists a point cc between x1x_1 and x2x_2 such that the relationship between the function values and the derivative is given by:

f(x2)−f(x1)=f′(c)(x2−x1)f(x_2) - f(x_1) = f'(c)(x_2 - x_1)

Given that f′(x)>0f'(x) > 0 for all x∈(a,b)x \in (a, b), it follows that f′(c)>0f'(c) > 0. Also, since x1<x2x_1 < x_2, the term (x2−x1)(x_2 - x_1) is strictly positive.

Therefore, the product f′(c)(x2−x1)f'(c)(x_2 - x_1) is greater than zero, which implies:

f(x2)−f(x1)>0f(x_2) - f(x_1) > 0

or equivalently,

f(x2)>f(x1)f(x_2) > f(x_1)

Thus, for any x1,x2∈[a,b]x_1, x_2 \in [a, b], if x1<x2x_1 < x_2, then f(x1)<f(x2)f(x_1) < f(x_2).

∴ ff is an increasing function on (a,b)(a, b).

Exercise: Miscellaneous Exercise on Chapter 6 | Q: 13 | (Page 38)
For More Understanding

Explanation

The answer is derived directly from Proof (a) found in the provided context. It utilizes the Mean Value Theorem to establish the relationship between the derivative and the function values. Since the derivative is positive and the interval difference is positive, their product is positive, proving the function is increasing.

Solution Steps

  1. Step 1: Let x1,x2∈[a,b]x_1, x_2 \in [a, b] such that x1<x2x_1 < x_2.

  2. Step 2: Apply the Mean Value Theorem to get f(x2)−f(x1)=f′(c)(x2−x1)f(x_2) - f(x_1) = f'(c)(x_2 - x_1) for some cc between x1x_1 and x2x_2.

  3. Step 3: Use the given condition f′(c)>0f'(c) > 0 and the fact that x2−x1>0x_2 - x_1 > 0 to deduce f(x2)−f(x1)>0f(x_2) - f(x_1) > 0.

  4. Step 4: Conclude f(x2)>f(x1)f(x_2) > f(x_1), proving ff is an increasing function.

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