Proving that if ( f'(x) leq 0 ) for all ( x ) in ([a, b]), then ( f ) is strictly increasing

In this article, we will present a detailed proof to demonstrate that if the derivative of a function ( f(x) ) is always non-positive within a given interval ([a, b]), then the function ( f ) is strictly increasing. This proof relies heavily on the Mean Value Theorem (MVT) and Rolle's Theorem. Let's dive into the step-by-step reasoning.

Step-by-Step Outline of the Proof

Step 1: State the Mean Value Theorem

The Mean Value Theorem (MVT) states that if a function ( f ) is continuous on the closed interval ([a, b]) and differentiable on the open interval ((a, b)), then there exists at least one point ( c ) in ((a, b)) such that:

[f'(c) frac{f(b) - f(a)}{b - a}]

Step 2: Apply the Conditions

Given that ( f'(x) leq 0 ) for all ( x ) in the interval ([a, b]), we can infer the following:

For any point ( c ) in ((a, b)), ( f'(c) leq 0 ). This implies that the slope of the tangent to the curve at any point ( c ) in ((a, b)) is non-positive or zero.

Step 3: Analyze the Implications

From the MVT, we have:

[f'(c) frac{f(b) - f(a)}{b - a}]

Since ( f'(c) leq 0 ), we can conclude that:

[frac{f(b) - f(a)}{b - a} leq 0]

Step 4: Derive the Conclusion

The inequality (frac{f(b) - f(a)}{b - a} leq 0) implies:

[f(b) - f(a) leq 0]

This means:

[f(b) leq f(a)]

Step 5: Generalize for Any Two Points

For any two points ( x_1, x_2 in [a, b] ) with ( x_1 leq x_2 ), we can apply the MVT again. There exists a point ( c ) in ((x_1, x_2)) such that:

[f'(c) frac{f(x_2) - f(x_1)}{x_2 - x_1}]

Again, since ( f'(c) leq 0 ), we conclude that:

[frac{f(x_2) - f(x_1)}{x_2 - x_1} leq 0]

This implies:

[f(x_2) - f(x_1) leq 0 quad Rightarrow quad f(x_2) leq f(x_1)]

Rolle's Theorem and Its Applications

Rolle's Theorem

Rolle's Theorem states that if a differentiable function vanishes at two points of its domain of definition, its derivative must vanish somewhere in between.

Proof: A differentiable function is necessarily continuous, and a continuous function defined on a closed interval necessarily has a minimum and a maximum somewhere on it. If ( f(x) f(y) 0 ) at two points ( x ) and ( y ) of ([a, b]), then either the function is identically zero on ([x, y]) (in which case its derivative is zero at all points of that interval) or else it has a positive maximum and/or a negative minimum at some point ( z ) strictly between ( x ) and ( y ). In that case, its derivative at ( z ) must be 0 since if ( z ) is a maximum, ( f(w) - f(z) leq 0 ) for all nearby points ( w ), so the ratio ( frac{f(w) - f(z)}{w - z} ) will be ( 0 ) if ( w eq z ) and ( 0 ) is its only possible common limit as ( w ) tends to ( z ). If ( z ) is a minimum, a similar reasoning with all inequalities reversed holds true.

Corollary: Between Any Two Points

Between any two points ( w ) and ( z ) of the interval of definition of a differentiable function, there is a point ( t ) such that:

[frac{f(w) - f(z)}{w - z} f(t)]

Proof: Apply Rolle's theorem to the function ( f(x) - f(z) - frac{f(w) - f(z)}{w - z}(x - z) ), which vanishes at both ( z ) and ( w ).

Informal Explanation

Informally, Rolle's theorem says that a smooth curve that crosses the x-axis at two points must have a horizontal tangent somewhere in between, viz., at its maximum or minimum. The expression ( f(x) - f(z) - frac{f(w) - f(z)}{w - z}(x - z) ) is simply the difference between the graph of ( f ) and the chord through the points ( x, f(x) ) and ( y, f(y) ) of the graph.