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Chapter 5: Problem 36

Find all critical values, the largest open intervals on which \(f\) is increasing, the largest open intervals on which \(f\) is decreasing, and all relative maxima and minima. Sketch a rough graph of \(f\). In Exercises 37 through 42, assume that the constants \(a\) and \(b\) are positive. \(f(x)=e^{-x^{2}}\)

Short Answer

Expert verified
The critical point is at \(x = 0\), increasing on \((-\infty, 0)\), decreasing on \((0, \infty)\), with a relative maximum at \(x = 0\).

Step by step solution

01

Find the derivative of the function

To find the critical points, we first need the derivative of the function. Given the function \(f(x) = e^{-x^2}\), we will use the chain rule to differentiate it. The derivative \(f'(x)\) of \(e^{-x^2}\) is \(-2xe^{-x^2}\).
02

Identify critical points

Critical points occur where the derivative is zero or undefined. Here, \(f'(x) = -2xe^{-x^2}\). Since \(e^{-x^2}\) is never zero, we set \(-2x = 0\), resulting in \(x = 0\). Thus, the critical point is at \(x = 0\).
03

Determine intervals of increase/decrease

To classify the intervals, evaluate \(f'(x) = -2xe^{-x^2}\): - If \(x > 0\), then \(-2x < 0\), so \(f'(x) < 0\), meaning \(f(x)\) is decreasing on \((0, \infty)\). - If \(x < 0\), then \(-2x > 0\), so \(f'(x) > 0\), meaning \(f(x)\) is increasing on \((-\infty, 0)\).
04

Identify relative extrema

At \(x = 0\), the function transitions from increasing to decreasing, indicating a relative maximum at this point. To find the maximum value, calculate \(f(0) = e^0 = 1\). Thus, a relative maximum occurs at \(x = 0\) with a value of 1.

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Key Concepts

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

Critical Points
In calculus, critical points are crucial for understanding the behavior of a function. A critical point of a function occurs where its derivative is either zero or undefined. Calculating these points can tell us where a function could potentially have a local maximum or minimum. For example, in the given function \( f(x) = e^{-x^2} \), we first find its derivative \( f'(x) = -2xe^{-x^2} \).
Since the exponential function \( e^{-x^2} \) is never zero or undefined, we focus on \(-2x = 0\) to find the critical points. Solving it gives \( x = 0 \), meaning our critical point is at \( x = 0 \). This is where we begin our analysis of the function's behavior.
Derivative
The derivative is a fundamental concept in calculus. It measures the rate at which a function changes as its input changes. For the function \( f(x) = e^{-x^2} \), we apply the chain rule to find its derivative, resulting in \( f'(x) = -2xe^{-x^2} \).
This derivative helps us understand how \( f(x) \) behaves as \( x \) changes. The sign of the derivative indicates whether the function is increasing or decreasing at a particular point. Here, \( -2x \) leads the analysis, while \( e^{-x^2} > 0 \) for all \( x \). Hence, the derivative directly helps us explore the function's increasing or decreasing nature.
Intervals of Increase and Decrease
Determining the intervals on which a function increases or decreases involves examining the sign of its derivative. For \( f(x) = e^{-x^2} \), the derivative is \( f'(x) = -2xe^{-x^2} \).
  • If \( x > 0 \), then \( -2x < 0 \), meaning \( f'(x) < 0 \). Consequently, the function is decreasing on the interval \((0, \infty)\).
  • If \( x < 0 \), then \( -2x > 0 \), resulting in \( f'(x) > 0 \). Therefore, the function is increasing on the interval \((-\infty, 0)\).
These intervals provide insight into the function's behavior, essential for sketching graphs and understanding real-world applications.
Relative Maxima and Minima
A relative maximum or minimum is a point where a function reaches a peak or dip locally. In our example, the critical point \( x = 0 \) determines a relative maximum.
At this point, the function transitions from increasing to decreasing. To confirm, we evaluate \( f(0) = e^0 = 1 \). Therefore, the relative maximum occurs at \( x = 0 \) with a value of 1.
Identifying such points helps in analyzing graphical properties and optimizing real-world situations, showing where the function behaves optimally in a local sense.

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Most popular questions from this chapter

a. Suppose that the function \(y=f^{\prime}(x)\) is decreasing on the interval (0,2) and \(f^{\prime}(1)=0 .\) What can you say about a possible relative extremum of \(f(x)\) at \(x=1 ?\) Justify your answer. b. Suppose that the function \(y=g^{\prime}(x)\) is increasing on the interval (0,2) and \(g^{\prime}(1)=0 .\) What can you say about a possible relative extremum of \(g(x)\) at \(x=1\) ? Justify your answer.

The price of a commodity is given as a function of the demand \(x\). Use implicit differentiation to find \(\frac{d x}{d p}\) for the indicated \(x\). $$ p=-3 x+20, x=5 $$

The price of a commodity is given as a function of the demand \(x\). Use implicit differentiation to find \(\frac{d x}{d p}\) for the indicated \(x\). $$ p=\sqrt{25-x^{2}}, x=4 $$

Agriculture Feinerman and colleagues \(^{15}\) created a mathematical model that related the yield response of corn to nitrogen fertilizer was given by the equation \(y=-5.119+\) \(0.099 x-0.004 x^{1.5},\) where \(y\) is the yield measured in tons per hectare and \(x\) is in kilograms per hectare. Use calculus to find the value of \(x\) that maximizes yield.

Biological Control Lysyk \(^{86}\) studied the effect of temperature on various life history parameters of a parasitic wasp for the purposes of pest management. He collected data given in the following table that related temperature to the proportion that survived. \begin{tabular}{|l|ccccc|} \hline Temperature \(\left({ }^{\circ} \mathrm{C}\right)\) & 15 & 20 & 25 & 30 & 33 \\ \hline Proportion Survival & 0.31 & 0.87 & 0.96 & 0.94 & 0.74 \\ \hline \end{tabular} a. Find the best-fitting quadratic (as the researchers did). b. Find the temperature at which survival was maximum.
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