Problem 17 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=\left\\{\begin{array}{ll}x^{3}, & x \leq 1 \\ k x, & x>1\end{array}\right.\) (a) What values of \(k\) makes \(f(x)\) a continuous function? (b) If \(k\) is chosen so that \(f\) is continuous at \(x=1\), is \(f\) differentiable there?

Step-by-Step Solution

Answer

The function f(x) is continuous for k=1 and is not differentiable at x = 1.

Step 1 - Check Continuity

For the function f(x) to be continuous at \(x = 1\), it must satisfy the three conditions of continuity: \n1. Limit of f(x) as x approaches 1 from the left \(\lim_{{x \to 1^-}} f(x)\) exists. \n2. Limit of f(x) as x approaches 1 from the right \(\lim_{{x \to 1^+}} f(x)\) exists. \n3. Both these limits are equal to the value of the function at x=1, i.e., f(1). \n\nNow, calculate the limits and the function value at x = 1. Since f(x) = \(x^3\) when \(x \leq 1\), the limit of f(x) as x approaches 1 from the left and the function value at x=1 are both equal to \(1^3 = 1\). So, \(\lim_{{x \to 1^-}} f(x) = f(1) = 1\). \n\nSince f(x) = \(kx\) when \(x > 1\), the limit of f(x) as x approaches 1 from the right is \(k \times 1 = k\). So, \(\lim_{{x \to 1^+}} f(x) = k\).

Step 2 - Find 'k' for Continuity

For the function f(x) to be continuous at \(x = 1\), the limit from the right must equal the limit from the left, as well as the function value at x=1. So, we must have \(1 = k\). Therefore, 'k' must be 1 for the function to be continuous.

Step 3 - Check Differentiability

For the function f(x) to be differentiable at \(x = 1\), again we need to calculate the limit of the gradient as x approaches 1 from the left and from the right, and these two limits should be equal. The derivative of \(x^3\) is \(3x^2\), so the limit of the gradient as x approaches 1 from the left is \(3 \times 1^2 = 3\). The derivative of \(kx\) is k (which we have found to be 1), so the limit of the gradient as x approaches 1 from the right is 1. Since these two limits are not equal, the function is not differentiable at \(x = 1\).

Key Concepts

Piecewise Functions

In mathematics, piecewise functions are those defined by different expressions based on the value of the input variable. These functions "switch" from one expression to another at specified values of the input. In the provided exercise, the function \(f(x)\) switches its form depending on whether \(x\) is less than or equal to 1, or greater than 1:

• For \(x \leq 1\), the function is defined as \(f(x) = x^3\). This is a polynomial function.
• For \(x > 1\), the function is \(f(x) = kx\). This represents a linear function.

Understanding piecewise functions is crucial as it helps you visualize how the function behaves differently on different intervals. They are frequently encountered in real-world applications where processes or phenomena change abruptly at certain thresholds.

Differentiability

Differentiability refers to the ability of a function to have a derivative at a given point. If a function is differentiable at a point, it means you can find a tangent line at that precise spot, indicating smoothness.
In the case of the given piecewise function \(f(x)\), we need to check differentiability at \(x=1\). While the function is defined smoothly within each interval (either \(x^3\) or \(kx\) with \(k=1\)), the transition point \(x=1\) is critical. The derivatives:

• For \(x \leq 1\), the derivative is \(3x^2\) which, at \(x=1\), evaluates to \(3\).
• For \(x > 1\) where \(k=1\), the derivative is simply \(k = 1\).

For differentiability, the left-hand derivative must equal the right-hand derivative at \(x=1\), which is not the case here. Because these derivatives do not match, \(f(x)\) is not differentiable at \(x=1\).

Limit Calculations

Limit calculations are essential for analyzing the behavior of functions as inputs approach certain values. Specifically, for piecewise functions, limits help us determine conditions for continuity and differentiability.
In the exercise, limit calculations ensure continuity of \(f(x)\) at \(x=1\):

• Calculate \(\lim_{{x \to 1^-}} f(x)\): Since \(f(x) = x^3\) for \(x \leq 1\), this limit equals \(1^3 = 1\).
• Calculate \(\lim_{{x \to 1^+}} f(x)\): Since \(f(x) = kx\) for \(x > 1\), this limit equals \(k \times 1 = k\).

For \(f(x)\) to be continuous, these must be equal and also match \(f(1)\): \(k\) must be \(1\). Through these limits, we concretely analyze the function's behavior at key transition points.