Question

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

Short 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-by-step solution

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 \((a, b)\).

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.

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 \( f'(c) = (f(x_2) - f(x_1)) / (x_2 - x_1) \).

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 \( 0 = (f(x_2) - f(x_1)) / (x_2 - x_1) \).

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 \((a, b)\).