Problem 76 - Calculus Single Variable

Question

Define \(f(x)=\sin (1 / x)\) for \(x\) in the interval (0,1) . Use the Mean Value Theorem to show that \(f^{\prime}\) takes on all real values.

Step-by-Step Solution

Answer

The derivative \(f'(x)\) takes all real values because \(\cos(1/x)\) oscillates and \(-1/x^2\) modifies its magnitude.

Step 1: Mean Value Theorem Understanding

The Mean Value Theorem (MVT) states that if a function is continuous on a 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: Verify Continuity and Differentiability of f(x)

The function \(f(x) = \sin(1/x)\) is continuous on \((0,1)\) as \(x \to 0\) from the positive side, and differentiable on the open interval \((0,1)\), since the sine function and the reciprocal \(1/x\) are both differentiable for \(x > 0\).

Step 3: Expression for f'(x)

To find \(f'(x)\), apply the chain rule to \(f(x) = \sin(1/x)\). Let \(u = 1/x\), then \(f(x) = \sin(u)\). The derivative \(f'(x) = \cos(u) \cdot \frac{du}{dx} = \cos(1/x) \cdot (-1/x^2)\). So, \(f'(x) = -\cos(1/x) / x^2\).

Step 4: Analyzing f'(x) Takes All Real Values

For every value of \(y\), there exists \(c \in (0,1)\) such that \(f'(c) = y\). As \(x\) approaches 0, \(1/x\) approaches infinity, and \(\cos(1/x)\) oscillates between \(-1\) and \(1\). Therefore, \(f'(x)\) can be arbitrarily large or small, depending on the exact value of \(x\) due to multiplication by \(-1/x^2\). Thus, \(f'(x)\) can take on all possible real values.

Key Concepts

Continuity

Continuity is an essential property for applying the Mean Value Theorem. A function is continuous on an interval if small changes in input only cause small changes in the output. For our function, \(f(x) = \sin(1/x)\), continuity must be examined in the open interval \((0,1)\). Here, as \(x\) approaches 0 from the positive side, the argument \(1/x\) heads toward infinity, however, the sine function remains bounded as it oscillates between -1 and 1.
This shows that small changes in \(x\) lead to small oscillations in \(f(x)\), ensuring continuity. Without continuity, the Mean Value Theorem could not guarantee any specific behavior within the interval.

Differentiability

Differentiability refers to whether a function has a defined derivative at every point in an interval. For \(f(x) = \sin(1/x)\), we need differentiability on the open interval \((0,1)\). The function \(\sin(u)\) where \(u = 1/x\), is differentiable, and so is \(1/x\).
By combining these, the chain rule helps us differentiate \(f(x)\).
If \(f(x)\) weren't differentiable, it would not be possible to use the derivative in the Mean Value Theorem. Differentiability ensures the function has a smooth, non-jumpy graph which can be vital for finding points like \(c\) where the derivative equals a specific slope, \(f'(c)\).

Chain Rule

The chain rule allows us to differentiate composite functions. For a function like \(f(x) = \sin(1/x)\), each part of the function \(\sin(u)\) and \(u=1/x\) require individual derivatives before applying the chain rule.
The derivative of \(\sin(u)\) is \(\cos(u)\), and the derivative of \(u = 1/x\) is \(-1/x^2\).
By the chain rule, \(f'(x) = \cos(u) \cdot \left(\frac{du}{dx}\right)\), leading to \(f'(x) = \cos(1/x) \cdot (-1/x^2)\).
This understanding is crucial since it provides the derivative necessary for exploring the Mean Value Theorem's implications.

Extreme Values

Extreme values refer to the maximum and minimum points a function can achieve. In the context of \(f'(x)\) for \(f(x) = \sin(1/x)\), while \(f(x)\) remains bounded, \(f'(x)\) does not have the same restriction. The derivative \(f'(x) = -\cos(1/x) / x^2\) can reach extreme values due to the oscillating nature of \(\cos(1/x)\).
As \(x\) approaches zero, \(\cos(1/x)\) fluctuates between -1 and 1, causing \(f'(x)\) to potentially take on any real value. This means \(f'(x)\) can reach extremely large positive or negative values, depending on the timing of \(x\).
These extreme values prove to be a validation for the Mean Value Theorem's role in ensuring the derivative captures every possible slope within a given interval.