Exam 2 Math 261 Applied Calculus April 21, 2020 Concepts 3 P

Exam 2math 261 Applied Calculus Iapril 21 2020concepts 3 Points Ea

Explain the difference between a relative and absolute maximum.

The first derivative of a function, roughly speaking, measures how fast a function is changing at a point x. That is, for a function f and point x, f′(x) tells us how fast the function is changing at the point x. What does f′′(x) tell us? (Hint: Think of the first derivative as the speed of a car, and the second derivative as the car’s acceleration.)

(True/False) If a function f has a critical point at x, then f(x) must either be a relative maximum or relative minimum.

(True/False) An inflection point of a function is a point across which the direction of concavity changes.

(True/False) If a function is increasing on an interval I, then f′(x) > 0 at every point x on the interval I.

(True/False) If a function is concave down on an interval I, then f′′(x)

For the following function, list all the critical points and state whether or not the critical point is a relative maximum or minimum (8 points): x1: relative min 2.

Sketch the graph of f(x) = x³ − 3x + 6 by following the procedure below. In each part, you must show all your work for full credit.

• Find f′(x) (5 points):
• Find the critical points of f. State them in the form (x, y). That is, for each critical point, find the corresponding y value and write the critical point as an ordered pair (5 points):
• Determine whether each critical point c is a relative maximum, relative minimum, or neither (5 points):
• Sketch the graph of f (showing concavity and inflection points is not necessary, but you must make clear intervals where f(x) is increasing/decreasing and what f(x) looks like at each critical point) (5 points):

Sketch a graph that matches the following description (4 points): G(x) has a negative first derivative over (−∞,−3) and a positive first derivative over (−3,∞).

Sketch a graph that matches the following description (4 points): G(x) has a positive first derivative over (−∞,−5), a negative first derivative over (−5, 0), and negative first but positive second derivatives over (0,∞).

Suppose f(x) = x³ − 12x. Sketch the function by following the procedure below. In each part, you must show all your work for full credit.

• Find f′(x) and f′′(x) (6 points):
• Find the critical points and classify each as producing a relative max, relative min, or neither. State each point in the form (x, y) (6 points):
• Find the inflection points. Make sure you show your justification for why the point is an inflection point (6 points):
• On which intervals is the function increasing, and which is it decreasing? (6 points):
• On which intervals is the function concave up, and which is it concave down? (6 points):
• Sketch the graph (you must show what the function looks like at each critical point, inflection point, and on the intervals you listed above) (6 points):

Sketch a graph that matches the following description (4 points): f is increasing and concave up on (−∞, 0), f is increasing and concave down on (0,∞).

Sketch a graph that matches the following description (6 points): f(0) = 5, f′(0) = 0, f′′(0) 0; f(4) = 3, f′(4) = 0, f′′(4)

Paper For Above instruction

The concepts of relative and absolute maxima are fundamental in calculus to analyze the behavior of functions. A relative maximum of a function is a point where the function's value is higher than all nearby points, meaning there exists some interval around that point where the function does not exceed it. Conversely, an absolute maximum is the highest point across the entire domain of the function. Understanding the difference helps in optimizing functions, especially in real-world applications like economics and engineering.

The first derivative f′(x) indicates the rate at which a function f(x) is changing at a specific point x. It essentially measures the slope of the tangent line to the graph of f at that point. The second derivative, f′′(x), offers insight into the curvature of the function—it measures the rate at which f′(x) changes as x varies. In physical terms, if the first derivative is interpreted as velocity, then the second derivative is acceleration, indicating how quickly the rate of change itself is changing.

Regarding the properties of derivatives and critical points, a critical point occurs where the derivative f′(x) equals zero or is undefined. However, not all critical points are maxima or minima; some could be points of inflection or saddle points. Therefore, the statement that every critical point must correspond to a maximum or minimum is False.

An inflection point is characterized by a change in the concavity of the function. When the graph transitions from concave up to concave down or vice versa, the point of transition is called an inflection point. This change indicates that the second derivative changes sign at that point, which is critical in understanding the shape of the graph and in identifying potential points of zero curvature.

If a function is increasing on an interval I, then the first derivative f′(x) > 0 for every x in I. This direct relationship between increasing behavior and positive derivative is a fundamental concept in calculus, providing a quick test to determine increasing intervals. Similarly, if a function is concave down, then the second derivative is less than zero, indicating the function's curvature opens downward.

For the critical points of the function f(x) = x³ − 3x + 6, first, compute f′(x) = 3x² − 3. Setting the derivative to zero, 3x² − 3 = 0, yields critical points at x = ±1. The corresponding y-values are f(1) = 1 − 3 + 6 = 4 and f(−1) = −1 + 3 + 6 = 8. To classify these points, evaluate the second derivative, f′′(x) = 6x, which gives f′′(1) = 6 > 0, indicating a local minimum at (1, 4), and f′′(−1) = -6

Graphically, the function increases on (−∞,−1), decreases from (−1,1), reaches its minimum at (1,4), then increases again on (1,∞). The function is concave up where the second derivative is positive, that is, for x > 0, and concave down where it is negative, for x

Likewise, for the function f(x) = x³ − 12x, critical points occur where f′(x) = 3x² −12 = 0, giving x= ±2. Corresponding y-values are f(−2) = (−2)³ − 12(−2) = -8 + 24 = 16, and f(2) = 8 − 24 = -16. Second derivatives, f′′(x) = 6x, determine concavity, being positive for x > 0 (concave up) and negative for x

Understanding these properties allows for accurate sketching of the graph and recognition of increasing/decreasing behavior, concavity, and inflection points. The detailed analysis of derivatives facilitates predicting the shape of the graph and interpreting real-world phenomena described by the functions.

References

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