Question

Suppose that \(f\) is continuous on \([a, b], f_{+}^{\prime}(a)<\mu

Short answer

Question: Prove that if a continuous function f on the interval [a, b] has one-sided derivatives f'₊(a) < μ < f'₋(b), then there exists a point c in (a, b) such that either f(c) - f(a) = μ(c - a) or f(c) - f(b) = μ(c - b).

Step-by-step solution

Define the function g

Introduce a new function \(g(x) = f(x) - \mu x\). Notice that \(g(x)\) is also continuous on \([a, b]\) since it is a combination of a continuous function \(f(x)\) and a linear function.

Apply the Mean Value Theorem to g

Since \(g\) is continuous on \([a, b]\), by the Mean Value Theorem we have that there exists a point \(c\) in \((a, b)\) such that $$ g'(c) = \frac{g(b) - g(a)}{b - a}. $$

Consider the possibilities

There are two possibilities that we need to consider: 1. If \(g'(c) = g'_{+}(a)\), then we have $$ \frac{g(b) - g(a)}{b - a} = g'_{+}(a). $$ But, \(g'_{+}(a) = f_{+}'(a) - \mu\) and hence $$ \frac{g(b) - g(a)}{b - a} = f'_{+}(a) - \mu. $$ 2. If \(g'(c) = g'_{-}(b)\), then we have $$ \frac{g(b) - g(a)}{b - a} = g'_{-}(b). $$ But, \(g'_{-}(b) = f_{-}'(b) - \mu\) and hence $$ \frac{g(b) - g(a)}{b - a} = f'_{-}(b) - \mu. $$

Solve for c in both possibilities

For possibility 1, substituting \(g(x) = f(x) - \mu x\), we get $$ \frac{f(b) - f(a) - \mu(b - a)}{b - a} = f'_{+}(a) - \mu. $$ Simplifying, we have $$ f(b) - f(a) = \mu(b - a), $$ which implies $$ f(c) - f(b) = \mu(c - b). $$ For possibility 2, substituting \(g(x) = f(x) - \mu x\), we get $$ \frac{f(b) - f(a) - \mu(b - a)}{b - a} = f'_{-}(b) - \mu. $$ Simplifying, we have $$ f(b) - f(a) = \mu(b - a) - (f'_{-}(b) - \mu)(b - a). $$ Noting that \((f'_{-}(b) - \mu)(b - a)> 0\) by our initial inequality, we can write it as $$ f(b) - f(a) < \mu(b - a), $$ which implies $$ f(c) - f(a) = \mu(c - a). $$ Thus, in either possibility, we have shown that either \(f(c) - f(a) = \mu(c - a)\) or \(f(c) - f(b) = \mu(c - b)\) for some \(c\) in \((a, b)\).

Key concepts

Continuity

Continuity is a fundamental concept in calculus that describes a function's smoothness across an interval. For a function to be continuous on a closed interval \([a, b]\), it must not have any breaks, jumps, or holes in that interval.
Mathematically, a function \(f(x)\) is continuous at a point \(c\) if:

• \(\lim_{{x \to c}} f(x) = f(c)\)
• The left-hand limit and right-hand limit at \(c\) are equal to \(f(c)\).

In the given problem, the function \(f\) is continuous on the interval \([a, b]\). This is crucial because continuity ensures that the function does not skip any values between \(f(a)\) and \(f(b)\). This property allows us to apply various theorems, such as the Mean Value Theorem (MVT), which demands continuity to provide its guarantees.

Differentiability

Differentiability refers to a function having a derivative at a given point within an interval. When a function is differentiable, it can be smoothly approximated by a tangent line at any given point.
A function is differentiable at a point \(c\) if the derivative \(f'(c)\) exists, which means:

• The function is smooth and there are no sharp edges or cusps at \(c\).
• The function is continuous around \(c\), implying differentiability requires continuity.

In our scenario, \(f\) is differentiable within \((a, b)\) because we are given the derivatives \(f_{+}^{\prime}(a)\) and \(f_{-}^{\prime}(b)\). This lets us use important calculus tools like the Mean Value Theorem, which provides insights about the overall behavior of \(f(x)\) between \(a\) and \(b\). By analyzing the derivatives, we can understand where the tangent line to the curve of \(f\) aligns with the slope \(\mu\).

Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that for any value between \(f(a)\) and \(f(b)\), there exists some point \(c\) in \((a, b)\) where the function \(f\) takes on that value. This is due to the function’s continuity on the interval.
Here’s a simple way to think about it:

• If \(f(a)\) is less than \(f(b)\), then every value between them must be hit at some point on the interval \([a, b]\).
• This ensures no value within the range is missed, thanks to the function being unbroken and continuous.

In the context of the exercise, ensuring continuity allows us to confidently apply concepts like the Mean Value Theorem. This is because the IVT assures that there will be points satisfying the required conditions between \(f(a)\) and \(f(b)\). The lack of equality between \((f(b) - f(a)) /(b-a)\) and \(\mu\) implies that there are values that \(f(x)\) can achieve in between, meeting the conditions required by the problem.