Problem 2 - Calculus

Question

In Problems \(1-20\), find the numbers \(c\) that make \(f(x)\) into (A) a continuous function and (B) a differentiable function. In one case \(f(x) \rightarrow f(a)\) at every point, in the other case \(\Delta f / \Delta x\) has a limit at every point. $$ f(x)=\left\\{\begin{array}{cl} \cos ^{3} x & x \neq \pi \\ c & x=\pi \end{array}\right. $$

Step-by-Step Solution

Answer

For continuity, \(c = -1\). This value also ensures differentiability at \(x = \pi\).

Step 1: Continuity at a point (A)

To find the number \(c\) that makes \(f(x)\) continuous at \(x = \pi\), we need the function value at \(x = \pi\), which is \(c\), and the limit of \(f(x)\) as \(x\) approaches \(\pi\) to be equal. Compute the limit: \(\lim_{x \to \pi} \cos^3 x = \cos^3(\pi) = (-1)^3 = -1\). Thus, \(f(\pi) = c = -1\).

Step 2: Differentiability at a point (B)

To ensure differentiability at \(x = \pi\), \(f(x)\) must not only be continuous but also have a continuous derivative at that point. The derivative of \(\cos^3(x)\) is \(-3\cos^2(x)\sin(x)\). Evaluating this at \(x = \pi\) gives \(-3(-1)^2\sin(\pi) = 0\). Thus, both the left-hand and right-hand derivatives at \(x = \pi\) are 0. With \(c = -1\), the function is differentiable at \(x = \pi\).

Key Concepts

Limit of a function

The concept of the limit of a function is foundational in calculus and helps us understand what value a function approaches as the input approaches a certain point. For the problem at hand, we are interested in the limit of \( f(x) = \cos^3 x \) as \( x \) approaches \( \pi \).
We concern ourselves with values of \( x \) that are close to \( \pi \) but not exactly \( \pi \) itself. The limit tells us what the function is doing at that point without direct substitution.
Given \( \lim_{x \to \pi} \cos^3 x \), we substitute \( \pi \) into the function: each \( \cos(\pi) = -1 \). Hence, \( \cos^3(\pi) = (-1)^3 = -1 \).
This means that as \( x \) gets closer to \( \pi \) on either side, \( f(x) \) approaches \(-1 \). This limit is key in determining the continuity and differentiability of the function at \( \pi \).

Continuous function

Continuity of a function at a point, let's say \( x = a \), occurs when the output does not "jump" as \( x \) approaches \( a \). This requires that the value of the function at \( a \) is the same as the limit as \( x \) approaches \( a \).
For \( f(x) \) to be continuous at \( x = \pi \), we match the function's value at \( \pi \), which is \( c \), with the limit we calculated earlier. Thus, \( f(\pi) = c \) should equal \(-1 \).
If these values are the same, the function is continuous. In our exercise, this is achieved by setting \( c = -1 \). Hence, there is no sudden change, or discontinuity, in the function at \( x = \pi \).
A continuous function allows us to plot a graph that is smooth without breaks or jumps, making it predictable and well-behaved in mathematical terms.

Differentiable function

A differentiable function near a point \( x = a \) means we can determine its slope or rate of change at that point. Differentiability implies continuity, but not vice versa.
For \( f(x) \) to be differentiable at \( x = \pi \), the function must have both left-hand and right-hand derivatives equal at \( \pi \). The derivative of our function, \( \cos^3(x) \), is \(-3\cos^2(x)\sin(x)\).
When evaluating this at \( x = \pi \), we find \(-3(-1)^2\sin(\pi) = 0 \). The zero result signifies that the rate of change from both directions at \( x = \pi \) is the same, confirming differentiability.
By setting \( c = -1 \), not only is \( f(x) \) continuous, but it also smoothly passes through \( x = \pi \), confirmed by the derivative showing no abrupt change in slope. This means we can apply calculus tools reliably, such as finding maximums or minimums.