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Chapter 2: Problem 71

Assume that \(f\) is differentiable over \((-\infty, \infty) .\) Classify each of the following statements as either true or false. If a statement is false, explain why. The function \(f\) can have a point of inflection at a critical value.

Short Answer

Expert verified
True, a point of inflection can occur at a critical value if the concavity changes and \( f'(x) = 0 \).

Step by step solution

01

Understanding Concepts

A point of inflection occurs where a function changes its concavity from concave up to concave down, or vice versa. A critical value is where the derivative is zero or undefined. Understanding these definitions is key to determining if they can coincide.
02

Derivatives and Points of Inflection

For a point of inflection to occur, the second derivative of the function, denoted as \( f''(x) \), must change sign. This means \( f''(x_0) = 0 \) or \( f''(x_0) \) is undefined at the point of inflection \( x_0 \).
03

Relating to Critical Points

A critical value is found when \( f'(x) = 0 \) or is undefined. A point of inflection does not strictly need the first derivative to be zero; it requires the change in concavity, shown by the behavior of the second derivative.
04

Can Critical Points Be Points of Inflection?

Though usually distinct in nature, a function can have a point of inflection at a critical value if both conditions of a critical point (\( f'(x) = 0 \)) and a change in concavity are simultaneously satisfied.
05

Conclusion

The statement is true. A function can have a point of inflection at a critical value if the function's concavity changes and \( f'(x) = 0 \) at the same point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inflection Point
An inflection point can be a bit like a curve's way of switching gears. It's all about when a function shifts its behavior between curving upwards, known as concave up, and curving downwards, or concave down. This switch happens at an inflection point. Imagine driving on a hilly road. When the road changes from going uphill to downhill, that's similar to what a function does at an inflection point.
  • An inflection point occurs where the second derivative (\( f''(x) \)) of a function changes sign. This means moving from positive to negative or vice versa.
  • This doesn't necessarily happen only at points where the first derivative (\( f'(x) \)) equals zero; distinctively, it's all about the second derivative and concavity change.
Understanding this concept helps identify the flexibility of a curve's nature around certain points.
Critical Point
A critical point is where the function pauses or hesitates, showcasing a key change in its behavior. This often signifies a turning point for the function. Let's say you're hiking up a mountain. When you reach the peak, that's your critical point 鈥� you've stopped going upwards and might start descending.
  • A critical point occurs at places where the first derivative (\( f'(x) \)) of the function is zero or where it doesn't exist.
  • This point is significant because it can indicate local maxima or minima 鈥� where the function changes from increasing to decreasing or vice versa.
  • It helps in understanding places where the function's slope is flat or undefined, indicating potential very interesting behavior in the course of the graph.
These points are pivotal in sketching and analyzing the graph behavior of functions.
Concavity
Concavity speaks volumes about the direction a curve is facing. It describes whether a function鈥檚 curve is bending upwards or downwards. It's like the difference between a smile and a frown when you look at the graph.
  • If the function is concave up, it means the curve resembles a cup or a bowl that's open upwards, like a smile.
  • If it's concave down, it looks like an arch or a frown. This is determined by the sign of the second derivative (\( f''(x) \)).
  • Positive second derivative indicates concave up, while a negative second derivative points to concave down.
Grasping concavity is essential for understanding the overall shape and behavior of a graph over different intervals.
Second Derivative
The second derivative serves as an illuminating tool, especially when analyzing the concavity and behavior of a graph. While the first derivative gives us the slope or rate of change, the second derivative tells us how the rate itself is changing.
  • When you take the derivative of the function's derivative, you get the second derivative (\( f''(x) \)).
  • This derivative is integral in understanding not just 'where' a function flattens out (via the critical point), but also 'how' it behaves around those points, as seen in the concavity.
  • It's the main player for identifying inflection points, showing where a curve switches its concavity.
Interpreting the second derivative allows not just a glance at the slope, but a story about the curvature of the function's graph. It's a crucial player in graph analysis.

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Most popular questions from this chapter

How is the second derivative useful in finding the absolute extrema of a function?

Find the absolute extrema of each function, if they exist, over the indicated interval. Also indicate the \(x\) -value at which each extremum occurs. When no interval is specified, use the real numbers, \((-\infty, \infty)\). $$g(x)=\frac{1}{3} x^{3}+2 x^{2}+x ; \quad[-4,0]$$

Find the absolute maximum and minimum values of each function, and sketch the graph. $$h(x)=\left\\{\begin{array}{ll}1-x^{2}, & \text { for }-4 \leq x<0 \\\1-x, & \text { for } 0 \leq x<1 \\\x-1, & \text { for } 1 \leq x \leq 2\end{array}\right.$$

Body surface area. Certain chemotherapy dosages depend on a patient's surface area. According to the Mosteller model, $$ s=\frac{\sqrt{h w}}{60} $$ where \(h\) is the patient's height in centimeters, \(w\) is the patient's weight in kilograms, and \(S\) is the approximation of the patient's surface area in square meters. (Source: www.halls.md.) Assume that Kim's height is a constant \(165 \mathrm{~cm},\) but she is losing weight. If she loses \(2 \mathrm{~kg}\) per month, how fast is her surface area decreasing at the instant she weighs \(70 \mathrm{~kg}\) ?

Differentiate implicitly to find \(d^{2} y / d x^{2}\). $$ x y+x-2 y=4 $$
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