Problem 56 - Essential Calculus

Question

Prove that if \(f^{\prime}(x)=0\) for all \(x\) in an interval \((a, b),\) then \(f\) is constant on \((a, b)\).

Step-by-Step Solution

Answer

Since the derivative of the function \( f \) is zero for all \( x \) in the interval \((a, b)\), by the Mean Value Theorem, \( f \) must take on the same constant value at any two points in that interval. Therefore, \( f \) is a constant function in the interval \((a, b)\).

Step 1: Establish the assumption

Assume that \( f \) is a differentiable function on the interval \((a, b)\) and that its derivative \( f'(x) \) is equal to zero for all \( x \) in \((a, b)\).

Step 2: Choose two arbitrary points in the interval

Choose any two points \( x_1 \) and \( x_2 \) in the interval \((a, b)\) so that \( x_1 < x_2 \). Since the choice is arbitrary, any findings should apply to all pairs of points in the interval.

Step 3: Apply the Mean Value Theorem

According to the Mean Value Theorem, because \( f \) is continuous on the closed interval \([x_1, x_2]\) and differentiable on the open interval \((x_1, x_2)\), there exists some \( c \) in \( (x_1, x_2) \) such that \( f'(c) = (f(x_2) - f(x_1)) / (x_2 - x_1) \).

Step 4: Use the known derivative

We know from our assumption that \( f'(x) = 0 \) for all \( x \) in \((a, b)\). This means \( f'(c) = 0 \). So we have \( 0 = (f(x_2) - f(x_1)) / (x_2 - x_1) \).

Step 5: Solve for the function values

The above equation simplifies to \(f(x_2) - f(x_1) = 0\), or \( f(x_2) = f(x_1) \). This implies that the function \( f \) takes on the same value at any two points in the interval \((a, b)\). Therefore, \( f \) must be constant on \((a, b)\).

Key Concepts

Differentiable Function

A **differentiable function** is one that has a derivative at every point in its domain. This means you can find the derivative, or the rate of change, at each point. When a function is differentiable, it has a smooth, non-jagged curve.

• If a function is differentiable at a point, it must also be continuous at that point.
• Differentiability implies locally linear behavior—it can be closely approximated by a straight line at any small segment.

Differentiability is an important property in calculus because it allows us to use tools like the Mean Value Theorem. In proving that a function is constant on an interval, differentiability ensures that such tools work effectively.

Continuous Function

A **continuous function** is one where small changes in the input lead to small changes in the output. There are no sudden jumps or gaps in the graph of the function.

• For a function to be continuous on an interval, it must not break at any point within that interval.
• Continuity ensures that limits and certain theorems, like the Intermediate Value Theorem, can be applied.

In the context of the Mean Value Theorem, the function must be continuous on a closed interval to apply the theorem effectively. Without continuity, we couldn't guarantee the behavior necessary for the theorem to hold.

Derivatives

The **derivative** of a function measures how the function's value changes as its input changes. Formally, it's the limit of the average rate of change of the function over small intervals.

• The derivative at a specific point tells us the slope of the tangent line to the function at that point.
• If a derivative is zero over an entire interval, the function is constant over that interval.

In our exercise, assuming that the derivative is zero means that there is no change in the function value across the interval. This zero derivative is key to proving that the function remains constant, as there's no slope or change in height between any two points.