Problem 46 - Calculus Early Transcendentals

Question

Show that \(f(x)\) is continuous but not differentiable at the indicated point. Sketch the graph of \(f\) (a) \(f(x)=\sqrt[3]{x}, x=0\) (b) \(f(x)=\sqrt[3]{(x-2)^{2}}, x=2\)

Step-by-Step Solution

Answer

Both functions are continuous at the specified points, but neither is differentiable due to undefined or inconsistent slopes.

Step 1: Define Continuity

A function \(f(x)\) is continuous at \(x = c\) if \(\lim_{x \to c} f(x) = f(c)\). This means that as \(x\) approaches \(c\), \(f(x)\) approaches \(f(c)\). We will verify this for each part separately.

Step 2: Check Continuity for Part (a)

For \(f(x) = \sqrt[3]{x}\) at \(x = 0\), we identify \(f(0) = \sqrt[3]{0} = 0\). The limit \(\lim_{x \to 0} \sqrt[3]{x} = 0\) because the cube root function approaches zero as \(x\) approaches zero. Thus, \(f(x)\) is continuous at \(x = 0\).

Step 3: Check Differentiability for Part (a)

A function \(f(x)\) is differentiable at \(x = c\) if the derivative \(f'(c)\) exists. Consider \(f(x) = x^{1/3}\). The derivative is \(f'(x) = \frac{1}{3}x^{-2/3}\). At \(x = 0\), \(f'(x) = \frac{1}{3}x^{-2/3}\) becomes undefined because of division by zero, hence, \(f(x)\) is not differentiable at \(x = 0\).

Step 4: Sketch Graph for Part (a)

The graph of \(f(x) = \sqrt[3]{x}\) transitions smoothly through the origin. It is 'S-shaped' where it flattens at \(x = 0\) but continues through that point without a sharp corner.

Step 5: Check Continuity for Part (b)

For \(f(x) = \sqrt[3]{(x-2)^2}\) at \(x = 2\), we find \(f(2) = \sqrt[3]{(2-2)^2} = \sqrt[3]{0} = 0\). The limit \(\lim_{x \to 2} \sqrt[3]{(x-2)^2} = 0\) as the function values get closer to zero approaching \(x = 2\). Thus, \(f(x)\) is continuous at \(x = 2\).

Step 6: Check Differentiability for Part (b)

To ascertain differentiability, \((x-2)^2\) inside the cube root function causes the graph to flatten at \(x = 2\), which indicates a cusp. Thus, \(f(x)\) is not differentiable at \(x = 2\) due to this cusp, as directional derivatives from either side do not give a consistent slope.

Step 7: Sketch Graph for Part (b)

The graph of \(f(x) = \sqrt[3]{(x-2)^2}\) shows a cusp at \(x = 2\). It is continuous without jumps, but the slope abruptly changes direction at \(x = 2\), indicating non-differentiability.

Key Concepts

Continuous Functions

A continuous function is one that you can draw on a graph without lifting your pencil from the paper. This means there are no breaks, jumps, or holes in the graph of the function. Mathematically, a function \(f(x)\) is continuous at a certain point \(x = c\) if the limit of \(f(x)\) as \(x\) approaches \(c\) is equal to \(f(c)\).

• The function should not have any abrupt changes in its value.
• The graph should not have any disjointed sections.

For the exercise given, we examined continuity at specific points of two functions. For \(f(x) = \sqrt[3]{x}\) at \(x = 0\) and \(f(x) = \sqrt[3]{(x-2)^2}\) at \(x = 2\), both functions pass this test of continuity as the value of the function matches the limit at the specified points.

Differentiability

Differentiability refers to the ability to find a derivative for a function at a point. If a function is differentiable at a point, it means there is a well-defined tangent at that point. However, if there is any abrupt change in direction, such as a cusp or a corner, the function is not differentiable there.If we look at differentiability for \(f(x) = \sqrt[3]{x}\) at \(x = 0\):- Although the function is continuous, its derivative involves division by zero at this point making it undefined.For \(f(x) = \sqrt[3]{(x-2)^2}\) at \(x = 2\), though continuous, the graph features a cusp, causing the derivative to change direction suddenly and not having a consistent value across both sides.

Derivatives

A derivative represents how a function changes at any given point, essentially the function's slope at that point. It captures either a linear tangent line or fails to exist at non-differentiable points like cusps or corners.Mathematically, for a function \(f(x)\):

• The derivative \(f'(x)\) can be thought of as the slope of the tangent.
• If \(f'(x)\) is undefined, it often indicates a non-smooth graph at that point.

In the exercise, we found that:- For \(f(x) = x^{1/3}\), at \(x = 0\), the derivative \(f'(x) = \frac{1}{3}x^{-2/3}\) becomes undefined.- In \(f(x) = \sqrt[3]{(x-2)^2}\), at \(x = 2\), the abrupt change in slope results in undefined directional derivatives.

Mathematical Graphing

Graphing functions involves plotting points on a coordinate system to visually interpret how the function behaves. Key features like continuity, differentiability, and cusps are often visible.- Graphs help us see if there are any discontinuities.- They show points where slopes may sharply change, indicating lack of differentiability.

• For \(f(x) = \sqrt[3]{x}\), the graph gently transitions through origin, consistent with continuity.
• For \(f(x) = \sqrt[3]{(x-2)^2}\), it reveals a sharp point at \(x = 2\), a classic cusp.

By analyzing graphs, we can better understand each function’s behavior at specific points.

Limits in Calculus

Limits help determine the expected value of a function as it approaches a particular point. In calculus, limits are fundamental to defining both continuity and derivatives.They are expressed as \( \lim_{x \to c} f(x) \), representing the intended approach of the function as \(x\) nears \(c\).- The concept of a limit is used to verify the continuity of a function. If the limit equals the function's value at that point, the function is continuous there.- Limits also play a role in examining derivatives, especially when exploring points of non-differentiability, confirming whether the slope converges to a finite number.For our functions:

• The limit \( \lim_{x \to 0} \sqrt[3]{x} = 0 \) established continuity at \(x = 0\).
• Similarly, \( \lim_{x \to 2} \sqrt[3]{(x-2)^2} = 0 \) confirmed continuity at \(x = 2\), even though differentiability failed.