Problem 6 - 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. $$ f(x)=\left\\{\begin{array}{rl} x^{3} & x \neq c \\ -8 & x=c \end{array}\right. $$

Step-by-Step Solution

Answer

The number \( c \) that makes \( f(x) \) continuous and differentiable is \( -2 \).

Step 1: Understand the Function

The function is given as a piecewise function: \( f(x) = x^3 \) when \( x eq c \), and \( f(x) = -8 \) when \( x = c \). We need to find the value of \( c \) for which the function is continuous and differentiable.

Step 2: Determine Continuity

For a function to be continuous at \( x = c \), the limit from both sides should equal the function value at \( c \). That is, \( \lim_{{x \to c}} f(x) = f(c) \). Thus, \( \lim_{{x \to c}} x^3 = -8 \). This implies \( c^3 = -8 \).

Step 3: Solve for Continuity Condition

Solve the equation \( c^3 = -8 \) to find \( c \). The solution is obtained by taking the cube root on both sides, yielding \( c = -2 \).

Step 4: Check Differentiability

For differentiability at \( x = c \), the derivatives from both sides must be equal. Here, the derivative of \( x^3 \,\) is \( 3x^2 \,\). At \( c = -2 \), the left-hand and right-hand derivatives must be \( 3(-2)^2 = 12 \). Since this matches from both directions, the function is differentiable at \( c = -2 \).

Step 5: Verify the Solution

Substituting \( c = -2 \) into steps 2 and 4 satisfies both the conditions for continuity and differentiability because \( \lim_{{x \to -2}} x^3 = -8 \) and the derivative \( 3x^2 \) yields \( 12 \) from both sides, which are consistent.

Key Concepts

Differentiability

Differentiability is a concept in calculus that means a function has a derivative at a particular point. In simple terms, a function is differentiable at a point if it can be smoothly approximated by a straight line at that point. For the function in our exercise, differentiability at the point \(x = c\) requires that the derivatives from both sides of \(c\) agree.
The derivative of \(x^3\) is \(3x^2\). At \(c = -2\), the derivative is \(3(-2)^2 = 12\). This matching from both sides of \(c\) tells us the function's slope is consistent at \(c = -2\).

• It indicates a seamless or smooth transition at that point.
• The function can be graphed without sharp corners at \(x = c\).

As we solved for \(c = -2\), this was verified by ensuring that the left-hand and right-hand derivatives were the same, thus making the function differentiable at that point.

Piecewise Functions

Piecewise functions are mathematical functions defined by different expressions in different intervals of their domain. They are composed of multiple sub-functions, each applying to a certain part of the domain. The given function is piecewise, combining \(x^3\) and \(-8\) at \(x = c\).
Understanding piecewise functions means recognizing that the behavior of the function can change drastically at boundaries. One key point is to examine these boundaries to determine continuity and differentiability.

• Identify different pieces within the function.
• Apply the conditions for continuity and differentiability at the transition points.

This exercise challenges us to find the correct \(c\) that satisfies these conditions. By substituting \(c = -2\) into each piecewise part, we ensure continuity and differentiability at \(x = c\). This is a typical process with piecewise functions.

Limits

Limits are fundamental in calculus and are particularly important when dealing with continuity and differentiability. They help describe the value that a function approaches as the input approaches a particular point. In this exercise, we evaluated \(\lim_{{x \to c}} x^3\) and ensured it equaled the function value \(-8\) at \(x = c\).
Using limits involves checking the behavior of the function's left and right side as they meet at a specific point. This is crucial when confirming both the continuity, where \(\lim_{{x \to c}} f(x) = f(c)\), and differentiability.

• Calculate limits from both sides.
• Ensure that the left-hand and right-hand limits agree.

In our example, we set the limit \(c^3 = -8\) to find \(c = -2\). Evaluating both the limit and its agreement with the function value ensures a seamless transition.

Calculus Problems

Calculus problems often involve a series of steps to find and confirm solutions to given conditions related to functions, such as finding points of continuity and differentiability. This exercise asks for the value of \(c\) that makes \(f(x)\) both continuous and differentiable.
When tackling calculus problems, breaking them into smaller, manageable steps aids in understanding. This usually involves:

• Identifying the conditions that need to be met (continuity, differentiability).
• Applying specific methods like solving for equal limits or matching derivatives.
• Verifying the solution through substitution back into the original function.

In our particular problem, we've determined \(c = -2\) by ensuring our conditions meet at each stage, verifying with limits and derivatives. Understanding the full properties of what makes a function continuous or differentiable is crucial for accurately solving these kinds of calculus problems.