Problem 53 - Calculus Single Variable

Question

Evaluate \(f^{\prime}(0)\) if \(f(x)=x^{\alpha}\) where \(\alpha \geq 1\). Show that \(f(x)=\) \(x^{\alpha}\) is not differentiable at 0 if \(\alpha<1\)

Step-by-Step Solution

Answer

\(f'(0) = 0\) for \(\alpha \geq 1\). \(f(x)\) is not differentiable at 0 if \(\alpha < 1\).

Step 1: Use the Definition of Derivative

To evaluate the derivative of the function \(f(x) = x^{\alpha}\) at \(x = 0\), we start by using the definition of the derivative at a point:\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]Since \( f(x) = x^{\alpha} \), we have:\[ f(h) = h^{\alpha} \text{ and } f(0) = 0^{\alpha} = 0 \text{ for } \alpha \geq 1 \]Substituting these into the derivative definition gives:\[ f'(0) = \lim_{h \to 0} \frac{h^{\alpha} - 0}{h} \]

Step 2: Simplify the Expression

We simplify further as follows:\[ \frac{h^{\alpha}}{h} = h^{\alpha - 1} \]So the derivative becomes:\[ f'(0) = \lim_{h \to 0} h^{\alpha - 1} \]

Step 3: Evaluate the Limit for \(\alpha \geq 1\)

For \(\alpha \geq 1\), the exponent \(\alpha - 1 \geq 0\). Hence, as \(h \to 0\), \(h^{\alpha - 1} \to 0\). Thus:\[ f'(0) = 0 \text{ for } \alpha \geq 1 \]

Step 4: Consider the Case \(\alpha < 1\)

For \(\alpha < 1\), the exponent \(\alpha - 1 \) is negative. Thus, \(h^{\alpha - 1}\) approaches infinity as \(h \to 0\) because the expression includes a division by a small positive power of \(h\), rather than approaching zero. Therefore, \(\lim_{h \to 0} h^{\alpha - 1}\) is undefined or infinite. This indicates that \(f'(0)\) does not exist for \(\alpha < 1\).

Key Concepts

Derivatives

Derivatives are at the heart of calculus and provide us with the rate at which a function changes at any point. They are crucial for understanding how functions behave. When you find the derivative of a function at a particular point, you are essentially determining the slope of the tangent line to the function at that point. For example, the process we performed to find \( f'(0) \) for \( f(x) = x^{\alpha} \) involves evaluating the function's behavior as we examine smaller and smaller changes around \( x = 0 \).
This method is known as finding the derivative through limits. By assessing the change in \( f(x) \) with increasingly small increments, we are able to calculate the instantaneous rate of change at the desired point.

• Increasing the power \( \alpha \) tends to simplify finding the derivative because it keeps the expression finite.
• For \( \alpha \geq 1 \), the derivative exists because the expression simplifies without becoming infinite.

Differentiability

A function's differentiability at a point determines whether its derivative exists there. If a function is differentiable at a given point, it means that there is a unique tangent line to its curve at that point. Differentiability implies that the function is smooth and continuous, without any jumps or cusps at the point of interest. In our original exercise, we look at \( f(x) = x^{\alpha} \).

• \( f(x) \) is only differentiable at \( x = 0 \) when \( \alpha \geq 1 \), ensuring that the resulting function behaves well enough at zero.
• When \( \alpha < 1 \), the function becomes non-differentiable at \( x = 0 \), as the exponent results in a division that leads to infinity.

Checking for differentiability involves confirming that the derivative is both finite and distinct. If the derivative doesn't exist, the function isn't smooth at that point, and so it lacks a tangent.

Limits

The power of calculus largely lies in the concept of limits. A limit allows us to find what value a function approaches as the input becomes arbitrarily close to some point, even if the function itself isn't defined at that exact spot. In finding derivatives, we often employ limits to determine the behavior of a function as we zero in on a particular input value.
When we found \( f'(0) \), we used the limit definition of the derivative. We evaluated:

• \( \lim_{h \to 0} \frac{x^{\alpha} - 0}{x} = \lim_{h \to 0} h^{\alpha - 1} \)
• For \( \alpha \geq 1 \), this results in zero as the expression simplifies without diverging to infinity.
• If \( \alpha < 1 \), the limit becomes unbounded, signaling a problem with differentiability.

Limits help us manage behaviors, whether they result in clean, expected numbers, or in showing where calculations break down.

Power Functions

Power functions take the form \( f(x) = x^{\alpha} \), which are critical in both pure and applied mathematics. They have characteristics that vary greatly depending on the value of \( \alpha \). These functions introduce us to interesting behaviors based on their exponents.
For our exercise, we looked closely at the power function where \( \alpha \) controls the overall behavior of the function.

• When \( \alpha \geq 1 \), power functions are generally well-behaved and differentiable at zero.
• For \( \alpha < 1 \), however, a power function is not differentiable at zero since it tends towards infinity as checks for differentiability fail.

Power functions showcase how subtle changes in an exponent can lead to drastically different properties.