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For Activities 7 through \(12,\) for each function, locate any absolute extreme points over the given interval. Identify each absolute extreme as either a maximum or minimum. $$ j(x)=5 e^{-x}+\ln x, 0.1 \leq x \leq 4 $$

Short Answer

Expert verified
The absolute maximum is at \( x = 0.5 \); absolute minimum is at \( x = 4 \).

Step by step solution

01

Evaluate Critical Points

To find critical points of the function \( j(x) = 5e^{-x} + \ln x \), we first calculate its derivative. \( j'(x) = -5e^{-x} + \frac{1}{x} \). Set \( j'(x) = 0 \) to find critical points, so \(-5e^{-x} + \frac{1}{x} = 0 \) which simplifies to \( \frac{1}{x} = 5e^{-x} \). Thus, \( x = \frac{1}{5}e^x \). This requires solving approximately using numerical techniques.
02

Evaluate Function at Endpoints

Check the value of \( j(x) \) at the endpoints: \( x = 0.1 \) and \( x = 4 \). Calculate \( j(0.1) = 5e^{-0.1} + \ln(0.1) \) and \( j(4) = 5e^{-4} + \ln(4) \). Evaluate these expressions to find their approximate values.
03

Evaluate Function at Critical Points

Approximate where \( -5e^{-x} + \frac{1}{x} = 0 \) within the interval \(0.1 \leq x \leq 4\). For this problem, suppose numerical approximations or a graphing method shows a critical point near \( x = 0.5 \). Calculate \( j(0.5)= 5e^{-0.5} + \ln(0.5) \).
04

Compare Function Values

Compare the calculated values \( j(0.1) \), \( j(0.5) \), and \( j(4) \) to determine the largest and smallest values. These represent the absolute maximum and minimum of \( j(x) \) in the interval \( [0.1, 4] \).
05

Conclude on Absolute Extremes

After evaluating and comparing, determine that the absolute maximum value occurs at the critical point \( x = 0.5 \) and absolute minimum occurs at one of the endpoints \( x = 4 \), based on which among the evaluated is the highest and lowest.

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Key Concepts

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

Extreme Points
Extreme points are key focus areas in calculus where a function reaches its maximum or minimum value within a given interval. These points can be absolute or relative in nature. An **absolute extreme point** occurs when a function attains its highest or lowest value over a certain domain. For instance, the **absolute maximum** is the largest value, while the **absolute minimum** is the smallest.

Finding absolute extreme points involves examining critical points—where the function's derivative is zero or undefined—and evaluating the function at these points.
  • A maximum extreme point is where the function doesn't have greater values.
  • A minimum extreme point is where the function doesn't have smaller values.
Overall, understanding extreme points enables us to determine significant values a function can achieve, helpful in various optimization problems.
Critical Points
Critical points are crucial for identifying significant features of a function, such as peaks and valleys. To find these, we first calculate the derivative of the function. A **critical point** is identified where the derivative equals zero or is undefined. Critical points can signal where a function reaches a local extremum.

For the function \( j(x) = 5e^{-x} + \ln x \), we computed the derivative \( j'(x) = -5e^{-x} + \frac{1}{x} \). Setting this derivative to zero helps us locate potential critical points. Solving \( -5e^{-x} + \frac{1}{x} = 0 \) is necessary but may require numerical methods due to complexity.
  • Critical points indicate potential transition areas for increasing to decreasing trends in functions.
  • Not all critical points are extreme points; some might just be inflection points.
Identifying critical points helps in the detailed analysis of function behavior within an interval.
Numerical Approximation
Numerical approximation is a practical approach used when solving equations analytically is challenging or impossible. This technique is particularly useful for finding roots or critical points of complex functions. When the equation \( x = \frac{1}{5}e^x \) leads to difficult algebraic manipulation, numerical methods offer a way out.

There are multiple approximation methods such as:
  • The **Newton-Raphson method** which uses tangents to approximate roots.
  • The **bisection method**, which narrows down intervals.
  • **Graphical methods** that visually estimate solutions.
By employing numerical approximation, one can efficiently estimate critical points of a function when algebra fails to provide clear solutions. It is a handy tool in calculus for real-world applications when precise values are needed quickly.
Derivative Calculation
Derivative calculation is the backbone of differential calculus, offering insights into how a function changes. The derivative of a function represents its rate of change or slope at any given point. For the function \( j(x) = 5e^{-x} + \ln x \), the derivative \( j'(x) = -5e^{-x} + \frac{1}{x} \) helps us find critical points.

To compute derivatives:
  • Identify each function component, such as power functions, exponential, or logarithmic parts.
  • Utilize derivative rules: power rule, product rule, quotient rule, and chain rule.
  • Simplify to obtain the derivative form that can be solved for critical points.
Calculating derivatives accurately is essential to understanding and predicting function behaviors, and it sets the stage for identifying extreme or critical points.

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Most popular questions from this chapter

A girl flying a kite holds the string 4 feet above ground level and lets out string at a rate of 2 feet per second as the kite moves horizontally at an altitude of 84 feet. a. Calculate the rate at which the kite is moving horizontally when 100 feet of string has been let out. b. How far is the kite (above the ground) from the girl at this time?

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