Problem 70 - Calculus - AP Edition

Question

You are asked in these exercises to determine whether a piecewise-defined function \(f\) is differentiable at a value \(x=x_{0}\) where \(f\) is defined by different formulas on different sides of \(x_{0}\). You may use without proof the following result, which is a consequence of the Mean-Value Theorem (discussed in Section 4.8 ). Theorem. Let \(f\) be continuous at \(x_{0}\) and suppose that \(\lim _{x \rightarrow x_{0}} f^{\prime}(x)\) exists. Then \(f\) is differentiable at \(x_{0}\) and \(f^{\prime}\left(x_{0}\right)=\lim _{x \rightarrow \infty} f^{\prime}(x)\) Let $$ f(x)=\left\\{\begin{array}{ll} x^{3}+\frac{1}{16}, & x<\frac{1}{2} \\ \frac{2}{4} x^{2}, & x \geq \frac{1}{2} \end{array}\right. $$ Determine whether \(f\) is differentiable at \(x=\frac{1}{2} .\) If so, find the value of the derivative there.

Step-by-Step Solution

Answer

The function is not differentiable at \( x=\frac{1}{2} \) since it is not continuous there.

Step 1: Check continuity at x₀

First verify that the piecewise function is continuous at \(x_0\): the left and right limits must equal \(f(x_0)\).

Step 2: Check left and right derivatives

Compute \(f'(x_0^-) = \lim_{h\to 0^-} \frac{f(x_0+h)-f(x_0)}{h}\) and \(f'(x_0^+) = \lim_{h\to 0^+} \frac{f(x_0+h)-f(x_0)}{h}\). The function is differentiable at \(x_0\) if and only if both limits exist and are equal.

Key Concepts

Differentiability of Piecewise Functions

Differentiability is a fundamental concept in calculus that describes whether a function has a derivative at a particular point. For a piecewise function, which is defined by different expressions over different intervals, determining differentiability at a boundary point can be intriguing.

For a piecewise function to be differentiable at a point \(x = x_0\), it needs to satisfy two conditions:

• The function must be continuous at \(x_0\).
• The derivative of the function must exist at \(x_0\).

Examine the behavior of the different sub-functions around \(x_0\). If they join smoothly without a jump or cusp, the function may be differentiable there.

In the problem at hand, the function is given as:

• \(f(x) = x^3 + \frac{1}{16}\) for \(x < \frac{1}{2}\).
• \(f(x) = \frac{1}{2}x^2\) for \(x \geq \frac{1}{2}\).

To confirm differentiability, check the continuity and see if derivatives from both sides converge to a single finite value. If not, the function isn't differentiable at the point in question.

Understanding Continuity in Functions

Continuity means that a function doesn't have any abrupt changes, jumps, or breaks at a particular point. It is vital for a function to be differentiable because differentiability implies continuity.

A function \(f\) is continuous at \(x = x_0\) if:

• The left-hand limit (as \(x\) approaches \(x_0\) from the left) exists.
• The right-hand limit (as \(x\) approaches \(x_0\) from the right) exists.
• Both limits are equal and the same as \(f(x_0)\).

In our exercise, to verify continuity at \(x = \frac{1}{2}\):

• Calculate the left-hand limit: \(\lim_{x \to \frac{1}{2}^-}(x^3 + \frac{1}{16}) = \frac{3}{16}\).
• Calculate the right-hand limit: \(\lim_{x \to \frac{1}{2}^+}(\frac{1}{2}x^2) = \frac{1}{8}\).

Since the two limits are not equal, the function \(f(x)\) is not continuous at \(x = \frac{1}{2}\). Without continuity, a function cannot be differentiable.

Grasping the Concept of Limits

Limits help us understand the behavior of a function as it approaches a specific point. They are essential in defining both continuity and differentiability.
For a function to be continuous at a point, its limit from the left side and the right side must be equal and also equal to the value of the function at that point. This ensures that there are no gaps or jumps in the function.
In our piecewise function, the jump in values as \(x\) approaches \(\frac{1}{2}\) from either side results in unequal limits, signifying the discontinuity. Thus, calculating these simple left and right limits can tell a lot about the smoothness of a function.

Application of the Mean Value Theorem

The Mean Value Theorem (MVT) is a key result in calculus that connects the values of a function to its derivative. The theorem requires a function to be both continuous on a closed interval, \([a, b]\), and differentiable on the open interval, \((a, b)\). It states that there's at least one point \(c\) in \((a, b)\) such that:

• \(f'(c) = \frac{f(b) - f(a)}{b - a}\).

For piecewise functions, MVT helps determine differentiability by verifying smooth transition across pieces.
In this exercise, even if \(f(x)\) was continuous at \(x = \frac{1}{2}\), you would need to ensure that the derivatives from each side converge to the same value to apply MVT effectively. Our example teaches that without continuity, we can't even step into checking differentiability since the prerequisites of the MVT are not met.