Differentiability and Continuity on an Interval

Each figure in Exercises 45–50 shows the graph of a function over a closed interval D. At what domain points does the function appear to be

c. neither continuous nor differentiable?

Give reasons for your answers.

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Examine the graph for any points where there is a jump, hole, or vertical asymptote, as these indicate discontinuity.

Identify the point at x = 0, where there is a hole in the graph, indicating the function is not continuous at this point.

Check for differentiability by looking for sharp corners or cusps in the graph, as these indicate non-differentiability.

Notice the sharp corner at x = -1, which suggests the function is not differentiable at this point.

Conclude that the function is neither continuous nor differentiable at x = 0 due to the hole, and not differentiable at x = -1 due to the sharp corner.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Continuity

A function is continuous at a point if the limit of the function as it approaches that point equals the function's value at that point. This means there are no breaks, jumps, or holes in the graph at that point. For a function to be continuous over an interval, it must be continuous at every point within that interval.

Differentiability

A function is differentiable at a point if it has a defined derivative at that point, which implies that the function must be continuous there. Differentiability also requires that the graph has a tangent line at that point, meaning it cannot have sharp corners or vertical tangents. If a function is not continuous at a point, it cannot be differentiable there.

Points of Non-Continuity and Non-Differentiability

Points where a function is neither continuous nor differentiable typically include locations where there are jumps, vertical tangents, or corners in the graph. In the provided graph, such points can be identified by observing where the function does not meet the criteria for continuity or where the slope abruptly changes, indicating a lack of differentiability.

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Analyzing Motion Using Graphs

[Technology Exercise] Exercises 31–34 give the position function s = f(t) of an object moving along the s-axis as a function of time t. Graph f together with the velocity function v(t) = ds/dt = f'(t) and the acceleration function a(t) = d²s/dt² = f''(t). Comment on the object’s behavior in relation to the signs and values of v and a. Include in your commentary such topics as the following:

c. When does it change direction?

s = t² - 3t + 2, 0 ≤ t ≤ 5

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Computer Explorations

Use a CAS to perform the following steps in Exercises 55–62.

b. Using implicit differentiation, find a formula for the derivative dy/dx and evaluate it at the given point P.

2y² + (xy)¹/³ = x² + 2, P(1,1)

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Textbook Question