Question

sign of \(f^{\prime}\) Assume that \(f\) is differentiable on \(a \leq x \leq b\) and that \(f(\)b\()<$$f$$(\)a\()\). Show that \(f^{\prime}\) is negative at some point between \(a\) and \(b\).

Short answer

By applying the Mean Value Theorem, it can concluded that the derivative of the function \(f'(c)\) is negative at some point within the open interval (a, b), given that f(b) < f(a).

Step-by-step solution

Apply Mean Value Theorem

The Mean Value Theorem (MVT) states that for a given planar arc between two endpoints, there is at least one point at which the tangent to the arc is parallel to the secant through its endpoints. In mathematical terms, if a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), there exists at least one number c in the interval (a, b) (that is a < c < b) such that: \(f'(c) = \frac{f(b) - f(a)}{b - a}\). According to MVT, let's find the value of \(f'(c)\) for some c in the open interval (a, b).

Evaluate the Derivative

Evaluating the derivative \(f'(c)\) would give us \(f'(c) = \frac{f(b) - f(a)}{b - a}\). Considering the conditions of the problem, f(b) < f(a), and it is clear that b > a, it follows that \(f'(c) < 0\(, for some c in the open interval (a, b).