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Chapter 10: Problem 1

Let \(f(x)=\frac{e^{x}}{x^{2}+1} .\) Find and classify the critical points.

Short Answer

Expert verified
The function \(f(x)=\frac{e^{x}}{x^{2}+1}\) has a critical point at \(x=1\). The point is neither a local maximum nor a local minimum as the second derivative test is inconclusive at \(x=1\).

Step by step solution

01

Find the derivative of the function

Applying the quotient rule of differentiation which states that \(\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{vu'-uv'}{v^2}\), where \(u = e^x\) and \(v = x^2 + 1\). Thus, \(u' = e^x\) and \(v' = 2x\). Therefore, \[f'(x) = \frac{(x^2+1)e^x - 2x e^x }{(x^2 + 1)^2}\]
02

Find the critical points

Now, set \(f'(x) = 0\) and solve for \(x\). This gives: \[0 = \frac{(x^2+1)e^x - 2x e^x }{(x^2 + 1)^2}\] Simplifying, the equation becomes: \[0 = (x^2 - 2x + 1)e^x\] So, \(x = 1\) is a critical point.
03

Classify the critical points

Finally, classify the critical point. Take the second derivative of the function and substitute \(x =1\) into the second derivative, \(f''(x)\). The critical point is a local minimum if \(f''(1) >0\), it is a local maximum if \(f''(1)<0\) and the test is inconclusive if \(f''(1) =0\).
04

Calculation of the second derivative

Finding the second derivative \(f''(x)\) requires applying the quotient rule twice. After some derivative calculations, we will have: \[f''(x)=\frac{(x^2-6x+1)e^x}{(x^2+1)^3}\]
05

Classification of the critical point

Substitute \(x = 1\) into \(f''(x)\) to retrieve \(f''(1)=0\). Thus, the second derivative test is inconclusive.

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Key Concepts

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

Understanding the Quotient Rule
When you need to differentiate a function that is expressed as a quotient of two functions, the quotient rule is your go-to tool. It makes the process straightforward by providing a specific formula:
\[\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}\]Here's how it works:
  • Identify the two parts of your function 鈥� the numerator (\(u\)) and the denominator (\(v\)).
  • Differentiate both parts: \(u'\) is the derivative of the numerator and \(v'\) is the derivative of the denominator.
  • Substitute into the formula to find the derivative of the quotient.
For the given function \(f(x) = \frac{e^x}{x^2 + 1}\), the numerator is \(e^x\) and the denominator is \(x^2 + 1\). By calculating \(u'\) as \(e^x\) and \(v'\) as \(2x\), you then apply these into the quotient rule formula.
This differentiation provides a useful expression for finding critical points, which are values of \(x\) where the derivative equals zero or is undefined.
Exploring the Second Derivative Test
The second derivative test determines if a critical point is a local maximum, minimum, or an inflection point. Here's a breakdown of how it works:
  • Calculate the second derivative of the function \(f(x)\).
  • Substitute the critical point found from the first derivative into the second derivative.
  • Assess the nature of the critical point:
    • If \(f''(x) > 0\), it's a local minimum.
    • If \(f''(x) < 0\), it's a local maximum.
    • If \(f''(x) = 0\), the test is inconclusive, indicating you might need another method to determine the behavior.
In the exercise, when \(f''(1) = 0\), the second derivative test does not provide a classification, suggesting that further investigation is needed to understand the point's characteristic.
The Basics of Differentiation in Calculus
Differentiation is a fundamental concept in calculus, used to analyze the rate at which things change. At its core, differentiation provides the derivative, which can inform us of the slope of a tangent to a curve at any given point.
With functions like \(f(x) = \frac{e^x}{x^2 + 1}\), differentiation allows us to solve for critical points 鈥� spots on the curve where the function reaches local maximums or minimums, or possibly changes concavity.
To find the critical points:
  • Compute the first derivative and set it to zero to locate critical points.
  • Use the second derivative to determine the nature of these points, as shown earlier.
Differentiation, therefore, is essential not only for classifying critical points but also for understanding the broader behavior of functions in various mathematical and real-world contexts.

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Most popular questions from this chapter

A tin can for garbanza beans is designed to be a cylinder with volume of 300 cubic centimeters. Denote the radius by \(r\) and the height by \(h\). The top and bottom are thicker than the sides; for the purposes of our model, we'll assume that they are made with a double thickness of aluminum. (a) Give an expression for the volume of the can. (b) Give an expression for the amount of material used. (Remember that the top and bottom of the can are two layers thick.) (c) Make the expression from part (b) into a function of \(r\) alone. (d) What radius minimizes the material used? (e) What are the dimensions of the 300 cubic-centimeter can that require the least amount of material?

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum? (c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value? \(f(x)=x^{3}-3 x+2\) on \((-\infty, \infty)\)

\(f(x)=\frac{1}{3} x^{3}-2 x-\frac{1}{x}\) (a) Find \(f^{\prime}\). (b) Find \(f^{\prime \prime}\). (c) Find all critical points. Which of these critical points are also stationary points? (d) Analyze the critical points. Are they local extrema? Global extrema? Points of inflection? (e) Is \(x=0\) the \(x\) -value of a critical point? Why or why not? (f) What is the absolute maximum value of the function? The absolute minimum value?

For each function: (a) Find all critical points on the specified interval. (b) Classify each critical point: Is it a local maximum, a local minimum, an absolute maximum, or an absolute minimum? (c) If the function attains an absolute maximum and/or minimum on the specified interval, what is the maximum and/or minimum value? \(f(x)=\frac{x^{3}}{3}+2 x+\frac{3}{x}\) on \([-3,0)\)

In Problems 1 through 12: (a) Find all critical points. (b) Find \(f^{\prime \prime} .\) Use the second derivative, wherever possible, to determine which critical points are local maxima and which are local minima. If the second derivative test fails or is inapplicable, explain why and use an alternative method for classifying the critical point. $$ f(x)=x^{6}+x^{4} $$
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